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|---|---|---|---|---|---|---|---|---|---|---|
1ktbe0f2q
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
The equation $$\arg \left( {{{z - 1} \over {z + 1}}} \right) = {\pi \over 4}$$ represents a circle with :
|
[{"identifier": "A", "content": "centre at (0, $$-$$1) and radius $$\\sqrt 2 $$"}, {"identifier": "B", "content": "centre at (0, 1) and radius $$\\sqrt 2 $$"}, {"identifier": "C", "content": "centre (0, 0) and radius $$\\sqrt 2 $$"}, {"identifier": "D", "content": "centre at (0, 1) and radius 2"}]
|
["B"]
| null |
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266796/exam_images/f29lrw36fjmxmetgizum.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265873/exam_images/uw851jxzxx4ohkmsafk9.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266519/exam_images/aowdrqzdcv1y1fdhfgqa.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Morning Shift Mathematics - Complex Numbers Question 83 English Explanation"></picture> <br><br>In $$\Delta$$OAC<br><br>$$\sin \left( {{\pi \over 4}} \right) = {1 \over {AC}}$$<br><br>$$ \Rightarrow AC = \sqrt 2 $$<br><br>Also, $$\tan {\pi \over 4} = {{OA} \over {OC}} = {1 \over {OC}}$$<br><br>$$\Rightarrow$$ OC = 1<br><br>$$\therefore$$ centre (0, 1); Radius = $$\sqrt 2 $$
|
mcq
|
jee-main-2021-online-26th-august-morning-shift
|
1ktirzvls
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
A point z moves in the complex plane such that $$\arg \left( {{{z - 2} \over {z + 2}}} \right) = {\pi \over 4}$$, then the minimum value of $${\left| {z - 9\sqrt 2 - 2i} \right|^2}$$ is equal to _______________.
|
[]
| null |
98
|
Let $$z = x + iy$$<br><br>$$\arg \left( {{{x - 2 + iy} \over {x + 2 + iy}}} \right) = {\pi \over 4}$$<br><br>$$\arg (x - 2 + iy) - \arg (x + 2 + iy) = {\pi \over 4}$$<br><br>$${\tan ^{ - 1}}\left( {{y \over {x - 2}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 2}}} \right) = {\pi \over 4}$$<br><br>$${{{y \over {x - 2}} - {y \over {x + 2}}} \over {1 + \left( {{y \over {x - 2}}} \right).\left( {{y \over {x + 2}}} \right)}} = \tan {\pi \over 4} = 1$$<br><br>$${{xy + 2y - xy + 2y} \over {{x^2} - 4 + {y^2}}} = 1$$<br><br>$$4y = {x^2} - 4 + {y^2}$$<br><br>$${x^2} + {y^2} - 4y - 4 = 0$$<br><br>locus is a circle with center (0, 2) & radius = $$2\sqrt 2 $$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266505/exam_images/ymchjtfklcmd2dq8kelg.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Mathematics - Complex Numbers Question 77 English Explanation"><br><br>min. value = $${(AP)^2} = {(OP - OA)^2}$$<br><br>$$ = {\left( {9\sqrt 2 - 2\sqrt 2 } \right)^2}$$<br><br>$$ = {\left( {7\sqrt 2 } \right)^2} = 98$$
|
integer
|
jee-main-2021-online-31st-august-morning-shift
|
1l54akox4
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>Let arg(z) represent the principal argument of the complex number z. Then, |z| = 3 and arg(z $$-$$ 1) $$-$$ arg(z + 1) = $${\pi \over 4}$$ intersect :</p>
|
[{"identifier": "A", "content": "exactly at one point."}, {"identifier": "B", "content": "exactly at two points."}, {"identifier": "C", "content": "nowhere."}, {"identifier": "D", "content": "at infinitely many points."}]
|
["C"]
| null |
<p>Let $$z = x + iy$$</p>
<p>$$\therefore$$ $$|z| = \sqrt {{x^2} + {y^2}} $$</p>
<p>Given, $$|z| = 3$$</p>
<p>$$\therefore$$ $$\sqrt {{x^2} + {y^2}} = 3$$</p>
<p>$$ \Rightarrow {x^2} + {y^2} = 9 = {3^2}$$</p>
<p>This represent a circle with center at (0, 0) and radius = 3</p>
<p>Now, given</p>
<p>$$\arg (z - 1) - \arg (z + 1) = {\pi \over 4}$$</p>
<p>$$ \Rightarrow \arg (x + iy - 1) - \arg (x + iy + 1) = {\pi \over 4}$$</p>
<p>$$ \Rightarrow \arg (x - 1 + iy) - \arg (x + 1 + iy) = {\pi \over 4}$$</p>
<p>$$ \Rightarrow {\tan ^{ - 1}}\left( {{y \over {x - 1}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 1}}} \right) = {\pi \over 4}$$</p>
<p>$$ \Rightarrow {\tan ^{ - 1}}\left( {{{{y \over {x - 1}} - {y \over {x + 1}}} \over {1 + {y \over {x - 1}} \times {y \over {x + 1}}}}} \right) = {\pi \over 4}$$</p>
<p>$$ \Rightarrow {\tan ^{ - 1}}\left( {{{{{xy + y - xy + y} \over {{x^2} - 1}}} \over {{{{x^2} - 1 + {y^2}} \over {{x^2} - 1}}}}} \right) = {\pi \over 4}$$</p>
<p>$$ \Rightarrow {\tan ^{ - 1}}\left( {{{xy + y - xy + y} \over {{x^2} - 1 + {y^2}}}} \right) = {\pi \over 4}$$</p>
<p>$$ \Rightarrow {{2y} \over {{x^2} - 1 + {y^2}}} = \tan \left( {{\pi \over 4}} \right)$$</p>
<p>$$ \Rightarrow 2y = {x^2} + {y^2} - 1$$</p>
<p>$$ \Rightarrow {x^2} + {y^2} - 2y - 1 = 0$$</p>
<p>$$ \Rightarrow {x^2} + {(y - 1)^2} = {(\sqrt 2 )^2}$$</p>
<p>This represent a circle with center at (0, 1) and radius $$\sqrt 2 $$.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5miv4n5/2aac6eff-97fb-4c96-bb71-32df8bb84dca/b7223400-0445-11ed-9d9a-21bdd0f00aa4/file-1l5miv4n6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5miv4n5/2aac6eff-97fb-4c96-bb71-32df8bb84dca/b7223400-0445-11ed-9d9a-21bdd0f00aa4/file-1l5miv4n6.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Complex Numbers Question 72 English Explanation"></p>
<p>From diagram you can see both the circles do not cut anywhere.</p>
|
mcq
|
jee-main-2022-online-29th-june-evening-shift
|
1l567v516
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>The number of elements in the set {z = a + ib $$\in$$ C : a, b $$\in$$ Z and 1 < | z $$-$$ 3 + 2i | < 4} is __________.</p>
|
[]
| null |
40
|
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8e8f52/d1800b00-4e42-4501-bb81-7122062be06f/25ad5460-8716-11ed-92bb-e57e64e1a06d/file-1lc8e8f53.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc8e8f52/d1800b00-4e42-4501-bb81-7122062be06f/25ad5460-8716-11ed-92bb-e57e64e1a06d/file-1lc8e8f53.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Morning Shift Mathematics - Complex Numbers Question 70 English Explanation"><br>
at line $y=-2$, we have $(5,-2)(6,-2)(1,-2)(0,-2)$ $\Rightarrow 4$ points <br><br>at line $y=-1$, we have $(4,-1)(5,-1)(6,-1)(2,-1)$ $(1,-1)(0,-1) \Rightarrow 6$ points
<br><br>
at line $y=0$, we have $(0,0)(1,0)(2,0)(3,0)(4,0)$ $(5,0)(6,0) \Rightarrow 7$ points
<br><br>
at line $y=1$, we have $(1,1),(2,1),(3,1),(4,1),(5,1)$ i.e. 5 points
<br><br>
symmetrically
<br><br>
at line $y=-5$, we have 5 points
<br><br>
at line $y=-4$, we have 7 points
<br><br>
at line $y=-3$, we have 6 points
<br><br>
So Total integral points $=2(5+7+6)+4$
<br><br>
$$
=40
$$
|
integer
|
jee-main-2022-online-28th-june-morning-shift
|
1l56q2sx0
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>The number of points of intersection of <br/><br/>$$|z - (4 + 3i)| = 2$$ and $$|z| + |z - 4| = 6$$, z $$\in$$ C, is :</p>
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}]
|
["C"]
| null |
<p>$${C_1}:|z - (4 + 3i)| = 2$$ and $${C_2}:|z| + |z - 4| = 6$$, $$z \in C$$</p>
<p>C<sub>1</sub> represents a circle with centre (4, 3) and radius 2 and C<sub>2</sub> represents a ellipse with focii at (0, 0) and (4, 0) and length of major axis = 6, and length of semi-major axis 2$$\sqrt5$$ and (4, 2) lies inside the both C<sub>1</sub> and C<sub>2</sub> and (4, 3) lies outside the C<sub>2</sub></p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5p8atn5/a1f6e8e9-ac22-432a-959c-96b00d255605/c11c6010-05c2-11ed-b1f1-b5d4135a474f/file-1l5p8atn6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5p8atn5/a1f6e8e9-ac22-432a-959c-96b00d255605/c11c6010-05c2-11ed-b1f1-b5d4135a474f/file-1l5p8atn6.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Evening Shift Mathematics - Complex Numbers Question 69 English Explanation"> </p>
<p> $$\therefore$$ number of intersection points = 2</p>
|
mcq
|
jee-main-2022-online-27th-june-evening-shift
|
1l6ggfmy6
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>Let O be the origin and A be the point $${z_1} = 1 + 2i$$. If B is the point $${z_2}$$, $${\mathop{\rm Re}\nolimits} ({z_2}) < 0$$, such that OAB is a right angled isosceles triangle with OB as hypotenuse, then which of the following is NOT true?</p>
|
[{"identifier": "A", "content": "$$\\arg {z_2} = \\pi - {\\tan ^{ - 1}}3$$"}, {"identifier": "B", "content": "$$\\arg ({z_1} - 2{z_2}) = - {\\tan ^{ - 1}}{4 \\over 3}$$"}, {"identifier": "C", "content": "$$|{z_2}| = \\sqrt {10} $$"}, {"identifier": "D", "content": "$$|2{z_1} - {z_2}| = 5$$"}]
|
["D"]
| null |
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbq29i/e583a584-7e22-4939-89d9-b1f38618bbd4/01c10960-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbq29j.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nbq29i/e583a584-7e22-4939-89d9-b1f38618bbd4/01c10960-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbq29j.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th July Morning Shift Mathematics - Complex Numbers Question 58 English Explanation"></p>
<p>$${{{z_2} - 0} \over {(1 + 2i) - 0}} = {{|OB|} \over {|OA|}}{e^{{{i\pi } \over 4}}}$$</p>
<p>$$ \Rightarrow {{{z_2}} \over {1 + 2i}} = \sqrt 2 {e^{{{i\pi } \over 4}}}$$</p>
<p>OR $${z_2} = (1 + 2i)(1 + i)$$</p>
<p>$$ = - 1 + 3i$$</p>
<p>$$\arg {z_2} = \pi - {\tan ^{ - 1}}3$$</p>
<p>$$|{z_2}| = \sqrt {10} $$</p>
<p>$${z_1} - 2{z_2} = (1 + 2i) + 2 - 6i = 3 - 4i$$</p>
<p>$$\arg ({z_1} - 2{z_2}) = - {\tan ^{ - 1}}{4 \over 3}$$</p>
<p>$$|2{z_1} - {z_2}| = |2 + 4i + 1 - 3i| = |3 + i| = \sqrt {10} $$</p>
|
mcq
|
jee-main-2022-online-26th-july-morning-shift
|
1l6rez1b9
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>Let $$\mathrm{S}=\{z=x+i y:|z-1+i| \geq|z|,|z|<2,|z+i|=|z-1|\}$$. Then the set of all values of $$x$$, for which $$w=2 x+i y \in \mathrm{S}$$ for some $$y \in \mathbb{R}$$, is :</p>
|
[{"identifier": "A", "content": "$$\\left(-\\sqrt{2}, \\frac{1}{2 \\sqrt{2}}\\right]$$"}, {"identifier": "B", "content": "$$\\left(-\\frac{1}{\\sqrt{2}}, \\frac{1}{4}\\right]$$"}, {"identifier": "C", "content": "$$\\left(-\\sqrt{2}, \\frac{1}{2}\\right]$$"}, {"identifier": "D", "content": "$$\\left(-\\frac{1}{\\sqrt{2}}, \\frac{1}{2 \\sqrt{2}}\\right]$$"}]
|
["B"]
| null |
$S:\{z=x+i y:|z-1+i| \geq|z|,|z|<2,|z-i|=|z-1|\}$
<br><br>
$$
|z-1+i| \geq|z|
$$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8c3waq/4ba6cc0d-ceab-4ef1-baed-c4899ebff5b2/d5a07220-870d-11ed-ae18-7336d1cc7e9d/file-1lc8c3war.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc8c3waq/4ba6cc0d-ceab-4ef1-baed-c4899ebff5b2/d5a07220-870d-11ed-ae18-7336d1cc7e9d/file-1lc8c3war.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Complex Numbers Question 49 English Explanation 1"><br>
$|z|<2$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8c4zsj/031ee17e-0a26-48dd-a170-9b19fe86b156/f422a830-870d-11ed-9cef-67b5ed764fcf/file-1lc8c4zsk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc8c4zsj/031ee17e-0a26-48dd-a170-9b19fe86b156/f422a830-870d-11ed-9cef-67b5ed764fcf/file-1lc8c4zsk.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Complex Numbers Question 49 English Explanation 2"><br>
$$
|z-i| = |z-1|
$$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8c6vi8/11bce565-65c8-4e70-b43c-71f126e96d89/28717710-870e-11ed-9cef-67b5ed764fcf/file-1lc8c6vi9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc8c6vi8/11bce565-65c8-4e70-b43c-71f126e96d89/28717710-870e-11ed-9cef-67b5ed764fcf/file-1lc8c6vi9.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh" alt="JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Complex Numbers Question 49 English Explanation 3"><br>
$\because \quad w \in S$ and $w=2 x+i y$
<br><br>
$2 x<\frac{1}{2} \quad\quad \therefore x<\frac{1}{4}$
<br><br>
$(2 x)^{2}+(-2 x)^{2}<4$
<br><br>
$4 x^{2}+4 x^{2}<4$
<br><br>
$x^{2}<\frac{1}{2} \Rightarrow x \in\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
<br><br>
$\therefore \quad x \in\left(-\frac{1}{2}, \frac{1}{4}\right]$
|
mcq
|
jee-main-2022-online-29th-july-evening-shift
|
1ldo6c5al
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>Let $$a,b$$ be two real numbers such that $$ab < 0$$. IF the complex number $$\frac{1+ai}{b+i}$$ is of unit modulus and $$a+ib$$ lies on the circle $$|z-1|=|2z|$$, then a possible value of $$\frac{1+[a]}{4b}$$, where $$[t]$$ is greatest integer function, is :</p>
|
[{"identifier": "A", "content": "$\\left(\\frac{1+\\sqrt{7}}{4}\\right)$"}, {"identifier": "B", "content": "$$\\frac{1}{2}$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$-$$1"}]
|
["C"]
| null |
$\begin{aligned} & \left|\frac{1+a i}{b+i}\right|=1 \\\\ & |1+i a|=|b+i| \\\\ & a^2+1=b^2+1 \Rightarrow \mathrm{a}=\pm \mathrm{b} \Rightarrow \mathrm{b}=-\mathrm{a} \quad \text { as } \mathrm{ab}<0 \\\\ & (\mathrm{a} +\mathrm{ib}) \text { lies on }|z-1|=|2 z| \\\\ & |a+i b-1|=2|a+i b| \\\\ & (a-1)^2+b^2=4\left(a^2+b^2\right) \\\\ & (a-1)^2=a^2=4\left(2 a^2\right) \\\\ & 1-2 a=6 a^2 \Rightarrow 6 a^2+2 a-1=0 \\\\ & a=\frac{-2 \pm \sqrt{28}}{12}=\frac{-1 \pm \sqrt{7}}{6} \\\\ & a=\frac{\sqrt{7}-1}{6} \text { and } b=\frac{1-\sqrt{7}}{6}\end{aligned}$
<br/><br/>$$
\begin{aligned}
& {[a]=0} \\\\
& \therefore \frac{1+[a]}{4 b}=\frac{6}{4(1-\sqrt{7})}=-\left(\frac{1+\sqrt{7}}{4}\right)
\end{aligned}
$$
<br/><br/>Similarly when $$a = {{ - 1 - \sqrt 7 } \over 6}\,\text { and }\,b = {{1 + \sqrt 7 } \over 6}$$
<br/><br/>then [$a$] = -1
<br/><br/>$$ \therefore $$ $${{1 + \left[ a \right]} \over {4b}} = {{1 - 1} \over {4 \times {{1 + \sqrt 7 } \over 6}}}$$ = 0
|
mcq
|
jee-main-2023-online-1st-february-evening-shift
|
1ldomvcuf
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>If the center and radius of the circle $$\left| {{{z - 2} \over {z - 3}}} \right| = 2$$ are respectively $$(\alpha,\beta)$$ and $$\gamma$$, then $$3(\alpha+\beta+\gamma)$$ is equal to :</p>
|
[{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "9"}]
|
["A"]
| null |
$$\left| {{{z - 2} \over {z - 3}}} \right| = 2$$
<br/><br/>$$
\begin{aligned}
& \sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2} \\\\
& \Rightarrow x^2+y^2-4 x+4=4 x^2+4 y^2-24 x+36 \\\\
& \Rightarrow 3 x^2+3 y^2-20 x+32=0 \\\\
& \Rightarrow x^2+y^2-\frac{20}{3} \mathrm{x}+\frac{32}{3}=0 \\\\
& \Rightarrow (\alpha, \beta)=\left(\frac{10}{3}, 0\right) \\\\
& \gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} \\\\
& 3(\alpha + \beta + \gamma)=3\left(\frac{10}{3}+\frac{2}{3}\right) \\\\
& =12
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-1st-february-morning-shift
|
1ldpswr3h
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>For all $$z \in C$$ on the curve $$C_{1}:|z|=4$$, let the locus of the point $$z+\frac{1}{z}$$ be the curve $$\mathrm{C}_{2}$$. Then :</p>
|
[{"identifier": "A", "content": "the curves $$C_{1}$$ and $$C_{2}$$ intersect at 4 points"}, {"identifier": "B", "content": "the curve $$C_{2}$$ lies inside $$C_{1}$$"}, {"identifier": "C", "content": "the curve $$C_{1}$$ lies inside $$C_{2}$$"}, {"identifier": "D", "content": "the curves $$C_{1}$$ and $$C_{2}$$ intersect at 2 points"}]
|
["A"]
| null |
Let $\mathrm{w}=\mathrm{z}+\frac{1}{\mathrm{z}}=4 \mathrm{e}^{\mathrm{i} \theta}+\frac{1}{4} \mathrm{e}^{-\mathrm{i} \theta}$
<br/><br/>$\Rightarrow \mathrm{w}=\frac{17}{4} \cos \theta+\mathrm{i} \frac{15}{4} \sin \theta$
<br/><br/>So locus of $w$ is ellipse $\frac{x^{2}}{\left(\frac{17}{4}\right)^{2}}+\frac{y^{2}}{\left(\frac{15}{4}\right)^{2}}=1$
<br/><br/>Locus of $\mathrm{z}$ is circle $\mathrm{x}^{2}+\mathrm{y}^{2}=16$
<br/><br/>So intersect at 4 points.
|
mcq
|
jee-main-2023-online-31st-january-morning-shift
|
1ldv236hn
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>Let $$\mathrm{z_1=2+3i}$$ and $$\mathrm{z_2=3+4i}$$. The set $$\mathrm{S = \left\{ {z \in \mathbb{C}:{{\left| {z - {z_1}} \right|}^2} - {{\left| {z - {z_2}} \right|}^2} = {{\left| {{z_1} - {z_2}} \right|}^2}} \right\}}$$ represents a</p>
|
[{"identifier": "A", "content": "hyperbola with the length of the transverse axis 7"}, {"identifier": "B", "content": "hyperbola with eccentricity 2"}, {"identifier": "C", "content": "straight line with the sum of its intercepts on the coordinate axes equals $$-18$$"}, {"identifier": "D", "content": "straight line with the sum of its intercepts on the coordinate axes equals $$14$$"}]
|
["D"]
| null |
$\left|z-z_{1}\right|^{2}-\left|z-z_{2}\right|^{2}=\left|z_{1}-z_{2}\right|^{2}$
<br/><br/>
$\Rightarrow(x-2)^{2}+(y-3)^{2}-(x-3)^{2}-(y-4)^{2}=1+1$
<br/><br/>
$\Rightarrow-4 x+4+9-6 y-9+6 x-16+8 y=2$
<br/><br/>
$\Rightarrow 2 x+2 y=14$
<br/><br/>
$\Rightarrow x+y=7$
|
mcq
|
jee-main-2023-online-25th-january-morning-shift
|
1lgq0ti75
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>Let $$w=z \bar{z}+k_{1} z+k_{2} i z+\lambda(1+i), k_{1}, k_{2} \in \mathbb{R}$$. Let $$\operatorname{Re}(w)=0$$ be the circle $$\mathrm{C}$$ of radius 1 in the first quadrant touching the line $$y=1$$ and the $$y$$-axis. If the curve $$\operatorname{Im}(w)=0$$ intersects $$\mathrm{C}$$ at $$\mathrm{A}$$ and $$\mathrm{B}$$, then $$30(A B)^{2}$$ is equal to __________</p>
|
[]
| null |
24
|
Given the expression for $w$ as :
<br/><br/>$$w = z\bar{z} + k_1z + k_2iz + \lambda(1+i), \quad \text{where } k_1, k_2 \in \mathbb{R}.$$
<br/><br/>1. If $w = x+iy$, we can separate this into the real and imaginary parts :
<br/><br/> The real part is: $$\text{Re}(w) = x^2 + y^2 + k_1x - k_2y + \lambda = 0.$$
<br/><br/> The imaginary part is: $$\text{Im}(w) = k_1y + k_2x + \lambda = 0.$$
<br/><br/>2. We know the circle C of radius 1 touches the line $y=1$ and the y-axis. The standard form of the equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. If the circle touches y-axis, then the x-coordinate of the center $h$ must be equal to the radius. Also, the circle touches the line $y=1$, so the y-coordinate of the center $k$ must be 1 unit above this line, hence $k=2$. So the circle's equation is $(x-1)^2 + (y-2)^2 = 1$.
<br/><br/>3. By comparing this with $\text{Re}(w) = 0$, we can see that $k_1=-2$, $k_2=4$, and $\lambda=4$.
<br/><br/>4. The equation for the curve $\text{Im}(w)=0$ becomes $-2y + 4x + 4 = 0$, which simplifies to $y = 2x + 2$.
<br/><br/>5. To find the intersection points A and B, we solve the system of equations consisting of the circle and the line. Substituting $y = 2x + 2$ into the equation of the circle gives $5x^2 - 2x = 0$, with solutions $x = 0$ and $x = 2/5$.
<br/><br/>6. Substituting $x = 0$ and $x = 2/5$ into $y = 2x + 2$ gives $y = 2$ and $y = 6/5$ respectively. So, the intersection points are A = $(0, 2)$ and B = $(2/5, 14/5)$.
<br/><br/>7. The distance between A and B is $AB = \sqrt{[(2/5)^2 + (4/5)^2]} = \sqrt{4/5}$.
<br/><br/>8. Finally, $30(AB)^2 = 30 \times (4/5) = 24$.
|
integer
|
jee-main-2023-online-13th-april-morning-shift
|
1lgrehu31
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>Let $$\mathrm{C}$$ be the circle in the complex plane with centre $$\mathrm{z}_{0}=\frac{1}{2}(1+3 i)$$ and radius $$r=1$$. Let $$\mathrm{z}_{1}=1+\mathrm{i}$$ and the complex number $$z_{2}$$ be outside the circle $$C$$ such that $$\left|z_{1}-z_{0}\right|\left|z_{2}-z_{0}\right|=1$$. If $$z_{0}, z_{1}$$ and $$z_{2}$$ are collinear, then the smaller value of $$\left|z_{2}\right|^{2}$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\frac{3}{2}$$"}, {"identifier": "B", "content": "$$\\frac{5}{2}$$"}, {"identifier": "C", "content": "$$\\frac{13}{2}$$"}, {"identifier": "D", "content": "$$\\frac{7}{2}$$"}]
|
["B"]
| null |
Given, $$
z_0=\frac{1+3 i}{2}, z_1=(1+i)
$$
<br/><br/>$$
\left|z_1-z_0\right|=\left|\frac{1-i}{2}\right|=\frac{1}{\sqrt{2}}
$$
<br/><br/>$$
\begin{aligned}
& \left|z_1-z_0\right|\left|z_2-z_0\right|=1 \\\\
& \Rightarrow \frac{1}{\sqrt{2}}\left|z_2-z_0\right|=1 \\\\
& \Rightarrow\left|z_2-z_0\right|=\sqrt{2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \frac{z_2-z_0}{z_1-z_0}=\frac{\left|z_2-z_0\right|}{\left|z_1-z_0\right|}( \pm 1)= \pm 2 \\\\
& z_2=z_0 \pm 2\left(z_1-z_0\right)
\end{aligned}
$$
<br/><br/>$$
z_2=2 z_1-z_0=\frac{3}{2}+\frac{1}{2} i \Rightarrow\left|z_2\right|^2=\frac{5}{2}
$$
<br/><br/>OR
<br/><br/>$$
z_2=3 z_0-2 z_1=\frac{-1}{2}+\frac{5}{2} i \Rightarrow\left|z_2\right|^2=\frac{13}{2}
$$
|
mcq
|
jee-main-2023-online-12th-april-morning-shift
|
1lh2z0axk
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>For $$\alpha, \beta, z \in \mathbb{C}$$ and $$\lambda > 1$$, if $$\sqrt{\lambda-1}$$ is the radius of the circle $$|z-\alpha|^{2}+|z-\beta|^{2}=2 \lambda$$, then $$|\alpha-\beta|$$ is equal to __________.</p>
|
[]
| null |
2
|
Given equation of circle,
<br/><br/>$$
\begin{aligned}
& \quad|z-\alpha|^2+|z-\beta|^2=2 \lambda \\\\
& \therefore 2 \lambda=|\alpha-\beta|^2 .........(i)
\end{aligned}
$$
<br/><br/>For circle,
<br/><br/>$$
\left|z-z_1\right|^2+\left|z-z_2\right|^2=\left|z_1-z_2\right|^2
$$
<br/><br/>$\begin{array}{lll}\text { Radius, } r=\frac{\left|z_1-z_2\right|}{2}=\frac{|\alpha-\beta|}{2}=\sqrt{\lambda-1} \\\\ \Rightarrow |\alpha-\beta|=2 \sqrt{\lambda-1} \\\\ \Rightarrow |\alpha-\beta|^2=4(\lambda-1) \\\\ \Rightarrow 2 \lambda=4(\lambda-1) [\because \text { Using Eq. (i)] } \\\\ \Rightarrow 2 \lambda=4 \\\\ \Rightarrow \lambda=2 \end{array}$
<br/><br/>$\begin{array}{ll}\Rightarrow |\alpha-\beta|^2=4 \\\\ \Rightarrow |\alpha-\beta|=2\end{array}$
|
integer
|
jee-main-2023-online-6th-april-evening-shift
|
lsaq2wp0
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
Let $\mathrm{P}=\{\mathrm{z} \in \mathbb{C}:|z+2-3 i| \leq 1\}$ and $\mathrm{Q}=\{\mathrm{z} \in \mathbb{C}: z(1+i)+\bar{z}(1-i) \leq-8\}$. Let in $\mathrm{P} \cap \mathrm{Q}$, $|z-3+2 i|$ be maximum and minimum at $z_1$ and $z_2$ respectively. If $\left|z_1\right|^2+2\left|z_2\right|^2=\alpha+\beta \sqrt{2}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ equals _____________.
|
[]
| null |
36
|
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsz1bfzj/da67c7ec-984a-45c9-af11-5958669a974a/42ad13f0-d280-11ee-9b77-fbceb54c8042/file-6y3zli1lsz1bfzk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsz1bfzj/da67c7ec-984a-45c9-af11-5958669a974a/42ad13f0-d280-11ee-9b77-fbceb54c8042/file-6y3zli1lsz1bfzk.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Morning Shift Mathematics - Complex Numbers Question 22 English Explanation">
<br><br>Clearly for the shaded region $z_1$ is the intersection of the circle and the line passing through $\mathrm{P}\left(\mathrm{L}_1\right)$ and $\mathrm{z}_2$ is intersection of line $\mathrm{L}_1 \& \mathrm{~L}_2$
<br><br>Circle : $(x+2)^2+(y-3)^2=1$
<br><br>$$
\begin{aligned}
& \mathrm{L}_1: \mathrm{x}+\mathrm{y}-1=0 \\\\
& \mathrm{~L}_2: \mathrm{x}-\mathrm{y}+4=0
\end{aligned}
$$
<br><br>On solving circle $\& \mathrm{~L}_1$ we get
<br><br>$$
\mathrm{z}_1:\left(-2-\frac{1}{\sqrt{2}}, 3+\frac{1}{\sqrt{2}}\right)
$$
<br><br>On solving $\mathrm{L}_1$ and $\mathrm{z}_2$ is intersection of line $\mathrm{L}_1 \& \mathrm{~L}_2$ we get $z_2:\left(\frac{-3}{2}, \frac{5}{2}\right)$
<br><br>$\begin{aligned} & \left|z_1\right|^2+2\left|z_2\right|^2=14+5 \sqrt{2}+17 \\\\ & =31+5 \sqrt{2}\end{aligned}$
<br><br>$\begin{array}{ll}\text { So } & \alpha=31 \\\\ & \beta=5 \\\\ & \alpha+\beta=36\end{array}$
|
integer
|
jee-main-2024-online-1st-february-morning-shift
|
lsbkg21j
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
If $S=\{z \in C:|z-i|=|z+i|=|z-1|\}$, then, $n(S)$ is :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "0"}]
|
["A"]
| null |
<p>$$|z-i|=|z+i|=|z-1|$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1dbkuq/8fa39b20-7b31-407e-a4e3-48710c870daf/c548b320-d3c8-11ee-a50b-bb659a2e1d74/file-1lt1dbkur.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1dbkuq/8fa39b20-7b31-407e-a4e3-48710c870daf/c548b320-d3c8-11ee-a50b-bb659a2e1d74/file-1lt1dbkur.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Morning Shift Mathematics - Complex Numbers Question 21 English Explanation"></p>
<p>$$\mathrm{ABC}$$ is a triangle. Hence its circum-centre will be the only point whose distance from A, B, C will be same.</p>
<p>So $$n(S)=1$$</p>
|
mcq
|
jee-main-2024-online-27th-january-morning-shift
|
jaoe38c1lscorvog
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>Let the complex numbers $$\alpha$$ and $$\frac{1}{\bar{\alpha}}$$ lie on the circles $$\left|z-z_0\right|^2=4$$ and $$\left|z-z_0\right|^2=16$$ respectively, where $$z_0=1+i$$. Then, the value of $$100|\alpha|^2$$ is __________.</p>
|
[]
| null |
20
|
<p>$$\begin{aligned}
& \left|z-z_0\right|^2=4 \\
& \Rightarrow\left(\alpha-z_0\right)\left(\bar{\alpha}-\bar{z}_0\right)=4 \\
& \Rightarrow \alpha \bar{\alpha}-\alpha \bar{z}_0-z_0 \bar{\alpha}+\left|z_0\right|^2=4 \\
& \Rightarrow|\alpha|^2-\alpha \bar{z}_0-z_0 \bar{\alpha}=2 \quad\text{......... (1)} \\
& \left|z-z_0\right|^2=16
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow\left(\frac{1}{\bar{\alpha}}-z_0\right)\left(\frac{1}{\alpha}-\bar{z}_0\right)=16 \\
& \Rightarrow\left(1-\bar{\alpha} z_0\right)\left(1-\alpha \bar{z}_0\right)=16|\alpha|^2 \\
& \Rightarrow 1-\bar{\alpha} z_0-\alpha \bar{z}_0+|\alpha|^2\left|z_0\right|^2=16|\alpha|^2 \\
& \Rightarrow 1-\bar{\alpha} z_0-\alpha \bar{z}_0=14|\alpha|^2 \quad\text{....... (2)}
\end{aligned}$$</p>
<p>From (1) and (2)</p>
<p>$$\begin{aligned}
& \Rightarrow 5|\alpha|^2=1 \\
& \Rightarrow 100|\alpha|^2=20
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-27th-january-evening-shift
|
luxwe3b8
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>Let $$z$$ be a complex number such that the real part of $$\frac{z-2 i}{z+2 i}$$ is zero. Then, the maximum value of $$|z-(6+8 i)|$$ is equal to</p>
|
[{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "$$\\infty$$"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& n=\frac{z-2 i}{z+2 i} \\
& \text { Let } z=x+i y \\
& n=\frac{x+(y-2) i}{x+(y+2) i} \times\left(\frac{x-(y+2) i}{x-(y+2) i}\right) \\
& \operatorname{Re}(n)=\frac{x^2+(y-2)(y+2)}{x^2+(y+2)^2}=0 \\
& \Rightarrow x^2+(y-2)(y+2)=0
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow x^2+y^2-4=0 \\
& \Rightarrow x^2+y^2=4 \\
& \text { also, }|z-(6+8 i)| \leq|z|+|-6-8 i| \\
& |z-(6+8 i)| \leq 2+10=12
\end{aligned}$$</p>
<p>Hence, Maximum value of $$|z-(6+8 i)|$$ is 12.</p>
|
mcq
|
jee-main-2024-online-9th-april-evening-shift
|
lv2er3qp
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>The area (in sq. units) of the region $$S=\{z \in \mathbb{C}:|z-1| \leq 2 ;(z+\bar{z})+i(z-\bar{z}) \leq 2, \operatorname{lm}(z) \geq 0\}$$ is</p>
|
[{"identifier": "A", "content": "$$\\frac{7 \\pi}{4}$$\n"}, {"identifier": "B", "content": "$$\\frac{3 \\pi}{2}$$\n"}, {"identifier": "C", "content": "$$\\frac{7 \\pi}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{17 \\pi}{8}$$"}]
|
["B"]
| null |
<p>$$|z-1| \leq 2 \quad \Rightarrow \quad(x-1)^2+y^2=4$$</p>
<p>$$\begin{aligned}
& z+\bar{z}+i(z-\bar{z}) \leq 2 \\
\Rightarrow \quad & x-y \leq 1 \\
& \operatorname{Im}(z) \geq 0 \\
\Rightarrow \quad & y \geq 0
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhf72y5/53205d83-bd1d-474f-a6fc-77a86f42ba34/4b0075d0-1801-11ef-b156-f754785ad3ce/file-1lwhf72y6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhf72y5/53205d83-bd1d-474f-a6fc-77a86f42ba34/4b0075d0-1801-11ef-b156-f754785ad3ce/file-1lwhf72y6.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Complex Numbers Question 7 English Explanation"></p>
<p>$$\begin{aligned}
& \text { Required area }=\left(\frac{\frac{3 \pi}{4}}{2 \pi}\right)(\pi)(2)^2 \\
& =\frac{3}{2} \pi
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-4th-april-evening-shift
|
lv9s1zxt
|
maths
|
complex-numbers
|
applications-of-complex-numbers-in-coordinate-geometry
|
<p>Let $$S_1=\{z \in \mathbf{C}:|z| \leq 5\}, S_2=\left\{z \in \mathbf{C}: \operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0\right\}$$ and $$S_3=\{z \in \mathbf{C}: \operatorname{Re}(z) \geq 0\}$$. Then the area of the region $$S_1 \cap S_2 \cap S_3$$ is :</p>
|
[{"identifier": "A", "content": "$$\\frac{125 \\pi}{24}$$\n"}, {"identifier": "B", "content": "$$\\frac{125 \\pi}{6}$$\n"}, {"identifier": "C", "content": "$$\\frac{125 \\pi}{12}$$\n"}, {"identifier": "D", "content": "$$\\frac{125 \\pi}{4}$$"}]
|
["C"]
| null |
<p>$$S_1=\{z \in C:|z| \leq 5\}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwekhn5c/6d7750d9-cc34-4e40-9c38-cefbaf7666fe/a4650700-166f-11ef-8416-25c08f86a011/file-1lwekhn5d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwekhn5c/6d7750d9-cc34-4e40-9c38-cefbaf7666fe/a4650700-166f-11ef-8416-25c08f86a011/file-1lwekhn5d.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Evening Shift Mathematics - Complex Numbers Question 2 English Explanation 1"></p>
<p>$$S_2=\operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0$$</p>
<p>Take $$z=x+i y$$</p>
<p>$$=\frac{x+i y+1-\sqrt{3} i}{1-\sqrt{3} i} \times \frac{1+\sqrt{3} i}{1+\sqrt{3} i}$$</p>
<p>$$\begin{aligned}
= & \frac{x+i y+1-\sqrt{3} i+\sqrt{3} i x-\sqrt{3} y+\sqrt{3} i+3}{1+3} \\
& =y+\sqrt{3} x \geq 0 \\
S_3= & x \geq 0
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lweku1ft/5285b6dd-899b-419d-8a11-3e28b70cd80a/fd24cb90-1670-11ef-8416-25c08f86a011/file-1lweku1fu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lweku1ft/5285b6dd-899b-419d-8a11-3e28b70cd80a/fd24cb90-1670-11ef-8416-25c08f86a011/file-1lweku1fu.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Evening Shift Mathematics - Complex Numbers Question 2 English Explanation 2"></p>
<p>$$\begin{aligned}
& y+\sqrt{3} x=0 \\
& y=-\sqrt{3} x \\
& \text { Slope }=-\sqrt{3} \\
& 90+\theta=120^{\circ} \\
& \theta=30^{\circ} \\
& 90-\theta=60^{\circ}
\end{aligned}$$</p>
<p>In first quadrant we have angle $$\frac{\pi}{2}$$</p>
<p>so, total angle $$90^{\circ}+60^{\circ}=150^{\circ}$$</p>
<p>So, area $$\frac{\pi r^2}{2 \pi} \times \frac{5 \pi}{6}=\frac{125 \pi}{12}$$</p>
|
mcq
|
jee-main-2024-online-5th-april-evening-shift
|
6qtmLybxMm3QFvTz
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
z and w are two nonzero complex numbers such that $$\,\left| z \right| = \left| w \right|$$ and Arg z + Arg w =$$\pi $$ then z equals
|
[{"identifier": "A", "content": "$$\\overline \\omega $$ "}, {"identifier": "B", "content": "$$ - \\overline \\omega $$ "}, {"identifier": "C", "content": "$$\\omega $$ "}, {"identifier": "D", "content": "$$ - \\omega $$ "}]
|
["B"]
| null |
Let $$\left| z \right| = \left| \omega \right| = r$$
<br><br>$$\therefore$$ $$z = r{e^{i\theta }},\omega = r{e^{i\phi }}$$
<br><br>where $$\,\,\theta + \phi = \pi .$$
<br><br>$$\therefore$$ $$z = r{e^{i\left( {\pi - \phi } \right)}} = r{e^{i\pi }}.$$
$${e^{ - i\phi }} = - r{e^{ - i\phi }} = - \overline \omega .$$
<br><br>[as $$\,\,\,\,\overline \omega = r{e^{ - i\phi }}$$ ]
|
mcq
|
aieee-2002
|
WOIIo6OF6hpOrGnb
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
If $$z$$ and $$\omega $$ are two non-zero complex numbers such that $$\left| {z\omega } \right| = 1$$ and $$Arg(z) - Arg(\omega ) = {\pi \over 2},$$ then $$\,\overline {z\,} \omega $$ is equal to
|
[{"identifier": "A", "content": "$$- i$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "- 1"}, {"identifier": "D", "content": "$$i$$"}]
|
["A"]
| null |
Given that,
<br>$$\left| {z\omega } \right| = 1$$
<br>$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 1
<br>$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$
<br><br>and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$
<br>$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$
<br><br>When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.
<br><br>So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.
<br><br>So we can assume,
<br>$${{z \over \omega }}$$ = $$ki$$
<br><br>$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$
<br><br>$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$
<br><br>$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$ [ as $$\left| i \right|$$ = 1 ]
<br><br>$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$$${1 \over {\left| \omega \right|}}$$ = $$k$$
<br><br>$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$ $$\left| z \right|$$ = $$k$$ [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]
<br><br>$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $$k$$
<br><br>$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$
<br><br>$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$
<br><br>As $${{z \over \omega }}$$ is imaginary so we can write,
<br><br>$${{z \over \omega }}$$ = $$ - {{\overline z } \over {\overline \omega }}$$
<br>[ When $$z$$ is imaginary then $$z$$ = $$-\overline z $$ ]
<br><br>$$ \Rightarrow $$ $$\overline z \omega $$ = $$ - z\overline \omega $$
<br><br>$$ \Rightarrow $$ $$\overline z \omega $$ = $$-{{z \over \omega }}$$.$$\overline \omega $$.$$\omega $$
<br><br>$$ \Rightarrow $$ $$\overline z \omega $$ = $$-{{z \over \omega }}$$.$${\left| \omega \right|^2}$$
<br><br>$$ \Rightarrow $$ $$\overline z \omega $$ = $$-ki$$.$${\left( {{1 \over {\sqrt k }}} \right)^2}$$
<br><br>$$ \Rightarrow $$ $$\overline z \omega $$ = $$-ki$$.$${1 \over k}$$
<br><br>$$ \Rightarrow $$ $$\overline z \omega $$ = $$-i$$
<br><br><b>Method 2 :</b>
<br><br>Given that,
<br>$$\left| {z\omega } \right| = 1$$
<br>$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 1
<br>$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$
<br><br>and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$
<br>$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$
<br><br>When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.
<br><br>So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.
<br><br>So we can assume,
<br>$${{z \over \omega }}$$ = $$ki$$
<br><br>$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$
<br><br>$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$
<br><br>$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$ [ as $$\left| i \right|$$ = 1 ]
<br><br>$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$$${1 \over {\left| \omega \right|}}$$ = $$k$$
<br><br>$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$ $$\left| z \right|$$ = $$k$$ [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]
<br><br>$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $$k$$
<br><br>$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$
<br><br>$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$
<br><br>(1) Magnitude of $$\overline z \omega $$ <br><br>= $$\left| {\overline z } \right|\left| \omega \right|$$ <br><br>= $$\left| z \right|\left| \omega \right|$$ [ as $$\left| z \right|$$ = $$\left| {\overline z } \right|$$ ]
<br><br> = $$\sqrt k $$.$${{1 \over {\sqrt k }}}$$
<br><br> = 1
<br><br>$$\therefore$$ The distance from the origin of $${\overline z \omega }$$ is 1.
<br><br>(2) Argument of $${\overline z \omega }$$ = $$Arg\left( {\overline z \omega } \right)$$
<br><br>= $$Arg\left( {\overline z } \right) + Arg\left( \omega \right)$$
<br><br>= $$-Arg\left( z \right) + Arg\left( \omega \right)$$
<br><br>= $$ - \left( {Arg\left( z \right) - Arg\left( \omega \right)} \right)$$
<br><br>= $$ - {\pi \over 2}$$
<br><br>$$\therefore$$ $${\overline z \omega }$$ is at (0, -1) on the negative side of imaginary axis and making an angle of $${\pi \over 2}$$ clockwise.
<img class="question-image" src="https://imagex.cdn.examgoal.net/7qhSHs1ANpaCp0dNu/geWSekyiopJP1qh86SBj3m7vTGItc/LQ9OeUltbbmWQsskFLgKQr/image.svg" loading="lazy" alt="AIEEE 2003 Mathematics - Complex Numbers Question 165 English Explanation">
<br><br>$$\therefore$$ $${\overline z \omega }$$ = 0 + (-1)$$ \times $$$$i$$ = $$-i$$
|
mcq
|
aieee-2003
|
9qdSyFDNne1uVHsv
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
Let z and w be complex numbers such that $$\overline z + i\overline w = 0$$ and arg zw = $$\pi $$. Then arg z equals :
|
[{"identifier": "A", "content": "$${{5\\pi } \\over 4}$$ "}, {"identifier": "B", "content": "$${{\\pi } \\over 2}$$ "}, {"identifier": "C", "content": "$${{3\\pi } \\over 4}$$"}, {"identifier": "D", "content": "$${{\\pi } \\over 4}$$ "}]
|
["C"]
| null |
Given $$\overline z + i\overline w = 0$$
<br><br>$$ \Rightarrow \overline z = - i\overline w $$
<br><br>$$ \Rightarrow \overline{\overline z} = - \overline {i\overline w } $$
<br><br>$$ \Rightarrow \overline{\overline z} = - \overline i \overline{\overline w} $$
<br><br>$$ \Rightarrow z = - \overline i w$$
<br><br>$$ \Rightarrow z = - \left( { - i} \right)w$$
<br><br>$$ \Rightarrow z = iw$$
<br><br>Now given that Arg(zw) = $$\pi $$
<br><br>$$ \Rightarrow $$ Arg(z$$ \times $$$${z \over i}$$) = $$\pi $$
<br><br>$$ \Rightarrow $$ Arg(z<sup>2</sup>) - Arg(i) = $$\pi $$
<br><br>$$ \Rightarrow $$ 2Arg(z) - $${\pi \over 2}$$ = $$\pi $$
<br><br>[ $$i$$ complex number represent (0, 1) point on imaginary axis and Arg($$i$$) means the angle made by the point (0, 1) with real axis which is $${\pi \over 2}$$]
<br><br>$$ \Rightarrow $$ 2Arg(z) = $${{3\pi } \over 2}$$
<br><br>$$ \Rightarrow $$ Arg(z) = $${{3\pi } \over 4}$$
|
mcq
|
aieee-2004
|
HB8aSQvg51aWaI5u
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
If $${z_1}$$ and $${z_2}$$ are two non-zero complex numbers such that $$\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$, then arg $${z_1}$$ - arg $${z_2}$$ is equal to :
|
[{"identifier": "A", "content": "$${\\pi \\over 2}\\,$$ "}, {"identifier": "B", "content": "$$ - \\pi $$ "}, {"identifier": "C", "content": "0 "}, {"identifier": "D", "content": "$${{ - \\pi } \\over 2}$$ "}]
|
["C"]
| null |
Given that, $$\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$
<br><br>$$\,\left| {{z_1} + {z_2}} \right|$$ is the vector sum of $${z_1}$$ and $${z_2}$$. So $$\,\left| {{z_1} + {z_2}} \right|$$ should be $$<$$ $$\left| {{z_1}} \right| + \left| {{z_2}} \right|$$ but here they are equal so $${z_1}$$ and $${z_2}$$ are collinear.
<br><br>S if $${z_1}$$ makes an angle $$\theta $$ with x axis then $${z_2}$$ will also make $$\theta $$ angle.
<br><br>$$\therefore$$ arg $${z_1}$$ - arg $${z_2}$$ = $$\theta $$ - $$\theta $$ = 0
|
mcq
|
aieee-2005
|
TErZStb9eMi3WgFR
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
If z is a complex number of unit modulus and argument $$\theta $$, then arg $$\left( {{{1 + z} \over {1 + \overline z }}} \right)$$ equals :
|
[{"identifier": "A", "content": "$$ - \\theta \\,\\,$$ "}, {"identifier": "B", "content": "$${\\pi \\over 2} - \\theta \\,$$ "}, {"identifier": "C", "content": "$$\\theta \\,$$ "}, {"identifier": "D", "content": "$$\\,\\pi - \\theta \\,\\,$$ "}]
|
["C"]
| null |
Given $$\,\,\,\,\left| z \right| = 1,\,\,\arg \,z = \theta $$
<br><br>As we know, $$\,\,\,\,\overrightarrow z = {1 \over z}$$
<br><br>$$\therefore$$ $$\,\,\,\,\arg \left( {{{1 + z} \over {1 + \overrightarrow z }}} \right) = \arg \left( {{{1 + z} \over {1 + {1 \over z}}}} \right)$$
<br><br>$$ = \arg \left( z \right) = \theta .$$
|
mcq
|
jee-main-2013-offline
|
8tnVLW8PvIETENfbmt3rsa0w2w9jx23bxo1
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = $${\pi \over 2}$$
, then :
|
[{"identifier": "A", "content": "$$z\\overline w = {{1 - i} \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$$\\overline z w = i$$"}, {"identifier": "C", "content": "$$z\\overline w = {{ - 1 + i} \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$$\\overline z w = -i$$"}]
|
["D"]
| null |
$$\left| {zw} \right| = 1$$<br><br>
$$ \Rightarrow $$ $$\left| z \right|\left| w \right| = 1$$<br><br>
Let $$w = {1 \over r}{e^{i\theta }}$$<br><br>
then z = $$r{e^{i\left( {\theta + {\pi \over 2}} \right)}}$$<br><br>
$$\overline z w = {e^{ - i\left( {\theta + {\pi \over 2}} \right)}}.{e^{i\theta }} = {e^{ - i(\pi /2)}} = - i$$<br><br>
$$z\overline w = {e^{i\left( {\theta + {\pi \over 2}} \right)}}.{e^{ - i\theta }} = {e^{i\pi /2}} = i$$
|
mcq
|
jee-main-2019-online-10th-april-evening-slot
|
n1yDVpjIDl2kKgsiz97k9k2k5fo1l04
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
If $${{3 + i\sin \theta } \over {4 - i\cos \theta }}$$, $$\theta $$ $$ \in $$ [0, 2$$\theta $$], is a real number, then an argument of <br/>sin$$\theta $$ + icos$$\theta $$ is :
|
[{"identifier": "A", "content": "$$\\pi - {\\tan ^{ - 1}}\\left( {{3 \\over 4}} \\right)$$"}, {"identifier": "B", "content": "$$ - {\\tan ^{ - 1}}\\left( {{3 \\over 4}} \\right)$$"}, {"identifier": "C", "content": "$${\\tan ^{ - 1}}\\left( {{4 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$\\pi - {\\tan ^{ - 1}}\\left( {{4 \\over 3}} \\right)$$"}]
|
["D"]
| null |
Let z = $${{3 + i\sin \theta } \over {4 - i\cos \theta }}$$
<br><br>= $${{3 + i\sin \theta } \over {4 - i\cos \theta }} \times {{\left( {4 + i\cos \theta } \right)} \over {\left( {4 + i\cos \theta } \right)}}$$
<br><br>= $${{\left( {12 - \sin \theta \cos \theta } \right) + i\left( {4\sin \theta + 3\cos \theta } \right)} \over {16 + {{\cos }^2}\theta }}$$
<br><br>As Z is purely real
<br><br>$$ \therefore $$ 4sin$$\theta $$ + 3cos$$\theta $$ = 0
<br><br>$$ \Rightarrow $$ tan $$\theta $$ = $$ - {3 \over 4}$$
<br><br>$$ \therefore $$ $$\theta $$ lies in the 2<sup>nd</sup> quadrant then
<br><br>arg(sinθ + icosθ) = $$\pi $$ + $${\tan ^{ - 1}}\left( {{{\cos \theta } \over {\sin \theta }}} \right)$$
<br><br>= $$\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)$$
|
mcq
|
jee-main-2020-online-7th-january-evening-slot
|
mK7IWzhYOPPZWqN3EDjgy2xukf45w8ky
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
If z<sub>1</sub>
, z<sub>2</sub>
are complex numbers such that
<br/>Re(z<sub>1</sub>) = |z<sub>1</sub> – 1|, Re(z<sub>2</sub>) = |z<sub>2</sub> – 1| , and
<br/>arg(z<sub>1</sub> - z<sub>2</sub>) = $${\pi \over 6}$$, then Im(z<sub>1</sub>
+ z<sub>2</sub>
) is equal to :
|
[{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${2\\sqrt 3 }$$"}]
|
["D"]
| null |
Let $${z_1} = {x_1} + i{y_1},\,{z_2} = {x_2} + i{y_2}$$
<br><br>Given Re(z<sub>1</sub>) = |z<sub>1</sub> – 1|
<br><br>$$ \therefore $$ x<sub>1</sub> = |(x<sub>1</sub> - 1) + iy<sub>1</sub>|
<br><br>$$ \Rightarrow $$ x<sub>1</sub> = $$\sqrt {{{\left( {{x_1} - 1} \right)}^2} + y_1^2} $$
<br><br>$$ \Rightarrow $$ $${x_1}^2 = {({x_1} - 1)^2} + {y_1}^2$$<br><br>$$ \Rightarrow {y_1}^2 - 2{x_1} + 1 = 0$$
<br><br>Also given Re(z<sub>2</sub>) = |z<sub>2</sub> – 1|
<br><br>$$ \therefore $$ x<sub>2</sub> = |(x<sub>2</sub> - 1) + iy<sub>2</sub>|
<br><br>$$ \Rightarrow $$ $${x_2}^2 = {({x^2} - 1)^2} + {y_2}^2$$<br><br>$$ \Rightarrow $$ $$y_2^2 - 2{x_2} - 1 = 0$$<br><br>Performing equation (1) - (2),<br><br>$$({y_1}^2 - {y_2}^2) + 2({x_2} - {x_1}) = 0$$<br><br>$$ \Rightarrow $$ $$({y_1} + {y_2})({y_1} - {y_2}) = 2({x_1} - {x_2})$$<br><br>$$ \Rightarrow $$ $${y_1} + {y_2} = 2\left( {{{{x_1} - {x_2}} \over {{y_1} - {y_2}}}} \right)$$<br><br>Now given, $$\arg ({z_1} - {z_2}) = {\pi \over 6}$$
<br><br>$$ \Rightarrow $$ $${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 6}$$<br><br>$$ \Rightarrow {{{y_1} - {y_2}} \over {{x_1} - {x_2}}} = {1 \over {\sqrt 3 }}$$<br><br>$$ \therefore $$ $${y_1} + {y_2} = 2\sqrt 3 $$
|
mcq
|
jee-main-2020-online-3rd-september-evening-slot
|
1krpubj40
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
If z and $$\omega$$ are two complex numbers such that $$\left| {z\omega } \right| = 1$$ and $$\arg (z) - \arg (\omega ) = {{3\pi } \over 2}$$, then $$\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$ is :<br/><br/>(Here arg(z) denotes the principal argument of complex number z)
|
[{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$$ - {{3\\pi } \\over 4}$$"}, {"identifier": "C", "content": "$$ - {\\pi \\over 4}$$"}, {"identifier": "D", "content": "$${{3\\pi } \\over 4}$$"}]
|
["B"]
| null |
As $$\left| {z\omega } \right| = 1$$<br><br>$$\Rightarrow$$ If $$\left| z \right| = r$$, then $$\left| \omega \right| = {1 \over r}$$<br><br>Let $$\arg (z) = \theta $$<br><br>$$\therefore$$ $$\arg (\omega ) = \left( {\theta - {{3\pi } \over 2}} \right)$$<br><br>So, $$z = r{e^{i\theta }}$$<br><br>$$ \Rightarrow \overline z = r{e^{i\theta }}$$<br><br>$$\omega = {1 \over r}{e^{i\left( {\theta - {{3\pi } \over 2}} \right)}}$$<br><br>Now, consider<br><br>$${{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }} = {{1 - 2{e^{i\left( { - {{3\pi } \over 2}} \right)}}} \over {1 + 3{e^{i\left( { - {{3\pi } \over 2}} \right)}}}} = \left( {{{1 - 2i} \over {1 + 3i}}} \right)$$<br><br>$$ = {{(1 - 2i)(1 - 3i)} \over {(1 + 3i)(1 - 3i)}} = - {1 \over 2}(1 + i)$$<br><br>$$\therefore$$ $$prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$<br><br>$$ = prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$<br><br>$$ = \left( { - {1 \over 2}(1 + i)} \right)$$<br><br>$$ = - \left( {\pi - {\pi \over 4}} \right) = {{ - 3\pi } \over 4}$$<br><br>So, option (2) is correct.
|
mcq
|
jee-main-2021-online-20th-july-morning-shift
|
1ktgosazo
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
Let z<sub>1</sub> and z<sub>2</sub> be two complex numbers such that $$\arg ({z_1} - {z_2}) = {\pi \over 4}$$ and z<sub>1</sub>, z<sub>2</sub> satisfy the equation | z $$-$$ 3 | = Re(z). Then the imaginary part of z<sub>1</sub> + z<sub>2</sub> is equal to ___________.
|
[]
| null |
6
|
Let z<sub>1</sub> = x<sub>1</sub> + iy ; z<sub>2</sub> = x<sub>2</sub> + iy<sub>2</sub><br><br>z<sub>1</sub> $$-$$ z<sub>2</sub> = (x<sub>1</sub> $$-$$ x<sub>2</sub>) + i(y<sub>1</sub> $$-$$ y<sub>2</sub>)<br><br>$$\therefore$$ $$\arg ({z_1} - {z_2}) = {\pi \over 4}$$ $$\Rightarrow$$ $${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$$<br><br>$${y_1} - {y_2} = {x_1} - {x_2}$$ ....... (1)<br><br>$$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$$ .... (2)<br><br>$$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$$ .... (3)<br><br>sub (2) & (3)<br><br>$${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$$<br><br>$$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$$<br><br>$$ = ({x_1} - {x_2})({x_1} + {x_2})$$<br><br>$${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$$
|
integer
|
jee-main-2021-online-27th-august-evening-shift
|
1l587df22
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
<p>Let $$A = \left\{ {z \in C:\left| {{{z + 1} \over {z - 1}}} \right| < 1} \right\}$$ and $$B = \left\{ {z \in C:\arg \left( {{{z - 1} \over {z + 1}}} \right) = {{2\pi } \over 3}} \right\}$$. Then A $$\cap$$ B is :</p>
|
[{"identifier": "A", "content": "a portion of a circle centred at $$\\left( {0, - {1 \\over {\\sqrt 3 }}} \\right)$$ that lies in the second and third quadrants only"}, {"identifier": "B", "content": "a portion of a circle centred at $$\\left( {0, - {1 \\over {\\sqrt 3 }}} \\right)$$ that lies in the second quadrant only"}, {"identifier": "C", "content": "an empty"}, {"identifier": "D", "content": "a portion of a circle of radius $${2 \\over {\\sqrt 3 }}$$ that lies in the third quadrant only"}]
|
["B"]
| null |
$$
\left|\frac{z+1}{z-1}\right|<1 \Rightarrow|z+1|<|z-1| \Rightarrow \operatorname{Re}(z)<0
$$<br><br>
and $\arg \left(\frac{z-1}{z+1}\right)=\frac{2 \pi}{3}$ is a part of circle as shown.<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc6ndrpb/ccdf08bc-7111-4568-85c5-b0eec7b3bc91/5b4a39e0-8620-11ed-91e0-bdec0331da70/file-1lc6ndrpc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc6ndrpb/ccdf08bc-7111-4568-85c5-b0eec7b3bc91/5b4a39e0-8620-11ed-91e0-bdec0331da70/file-1lc6ndrpc.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th June Morning Shift Mathematics - Complex Numbers Question 67 English Explanation">
|
mcq
|
jee-main-2022-online-26th-june-morning-shift
|
1l59jqnkg
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
<p>Let z<sub>1</sub> and z<sub>2</sub> be two complex numbers such that $${\overline z _1} = i{\overline z _2}$$ and $$\arg \left( {{{{z_1}} \over {{{\overline z }_2}}}} \right) = \pi $$. Then :</p>
|
[{"identifier": "A", "content": "$$\\arg {z_2} = {\\pi \\over 4}$$"}, {"identifier": "B", "content": "$$\\arg {z_2} = - {{3\\pi } \\over 4}$$"}, {"identifier": "C", "content": "$$\\arg {z_1} = {\\pi \\over 4}$$"}, {"identifier": "D", "content": "$$\\arg {z_1} = - {{3\\pi } \\over 4}$$"}]
|
["C"]
| null |
<p>$$\because$$ $${{{z_1}} \over {{z_2}}} = - i \Rightarrow {z_1} = - i{z_2}$$</p>
<p>$$ \Rightarrow \arg ({z_1}) = - {\pi \over 2} + \arg ({z_2})$$ ..... (i)</p>
<p>Also $$\arg ({z_1}) - \arg ({\overline z _2}) = \pi $$</p>
<p>$$ \Rightarrow \arg ({z_1}) + \arg ({z_2}) = \pi $$ ..... (ii)</p>
<p>From (i) and (ii), we get</p>
<p>$$\arg ({z_1}) = {\pi \over 4}$$ and $$\arg ({z_2}) = {{3\pi } \over 4}$$</p>
|
mcq
|
jee-main-2022-online-25th-june-evening-shift
|
1l5aj6yrj
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
<p>Let a circle C in complex plane pass through the points $${z_1} = 3 + 4i$$, $${z_2} = 4 + 3i$$ and $${z_3} = 5i$$. If $$z( \ne {z_1})$$ is a point on C such that the line through z and z<sub>1</sub> is perpendicular to the line through z<sub>2</sub> and z<sub>3</sub>, then $$arg(z)$$ is equal to :</p>
|
[{"identifier": "A", "content": "$${\\tan ^{ - 1}}\\left( {{2 \\over {\\sqrt 5 }}} \\right) - \\pi $$"}, {"identifier": "B", "content": "$${\\tan ^{ - 1}}\\left( {{{24} \\over 7}} \\right) - \\pi $$"}, {"identifier": "C", "content": "$${\\tan ^{ - 1}}\\left( 3 \\right) - \\pi $$"}, {"identifier": "D", "content": "$${\\tan ^{ - 1}}\\left( {{3 \\over 4}} \\right) - \\pi $$"}]
|
["B"]
| null |
<p>$${z_1} = 3 + 4i$$, $${z_2} = 4 + 3i$$ and $${z_3} = 5i$$</p>
<p>Clearly, $$C \equiv {x^2} + {y^2} = 25$$</p>
<p>Let $$z(x,y)$$</p>
<p>$$ \Rightarrow \left( {{{y - 4} \over {x - 3}}} \right)\left( {{2 \over { - 4}}} \right) = - 1$$</p>
<p>$$ \Rightarrow y = 2x - 2 \equiv L$$</p>
<p>$$\therefore$$ z is intersection of C & L</p>
<p>$$ \Rightarrow z \equiv \left( {{{ - 7} \over 5},{{ - 24} \over 5}} \right)$$</p>
<p>$$\therefore$$ $$Arg(z) = - \pi + {\tan ^{ - 1}}\left( {{{24} \over 7}} \right)$$</p>
|
mcq
|
jee-main-2022-online-25th-june-morning-shift
|
1ldr7wgx2
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
<p>Let $$z=1+i$$ and $$z_{1}=\frac{1+i \bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$$. Then $$\frac{12}{\pi} \arg \left(z_{1}\right)$$ is equal to __________.</p>
|
[]
| null |
9
|
<p>$$z = 1 + i$$</p>
<p>$${z_1} = {{1 + i\overline z } \over {\overline z (1 - z) + {1 \over z}}}$$</p>
<p>$$ = {{z(1 + i\overline z )} \over {|z{|^2}(1 - z) + 1}}$$</p>
<p>$$ = {{(1 + i)(1 + i(1 - i))} \over {2(1 - 1 - i) + 1}}$$</p>
<p>$${z_1} = 1 - i$$</p>
<p>$$\arg {z_1} = {\tan ^{ - 1}}\left( {{{ - 1} \over 1}} \right) = {{3\pi } \over 4}$$</p>
<p>$${{12} \over \pi }\arg ({z_1}) = {{3\pi } \over 4}\,.\,{{12} \over \pi } = 9$$</p>
|
integer
|
jee-main-2023-online-30th-january-morning-shift
|
1lgutz71b
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
<p>Let $$w_{1}$$ be the point obtained by the rotation of $$z_{1}=5+4 i$$ about the origin through a right angle in the anticlockwise direction, and $$w_{2}$$ be the point obtained by the rotation of $$z_{2}=3+5 i$$ about the origin through a right angle in the clockwise direction. Then the principal argument of $$w_{1}-w_{2}$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$-\\pi+\\tan ^{-1} \\frac{8}{9}$$"}, {"identifier": "B", "content": "$$-\\pi+\\tan ^{-1} \\frac{33}{5}$$"}, {"identifier": "C", "content": "$$\\pi-\\tan ^{-1} \\frac{8}{9}$$"}, {"identifier": "D", "content": "$$\\pi-\\tan ^{-1} \\frac{33}{5}$$"}]
|
["C"]
| null |
<p>To solve the problem, let's break it down step by step.</p>
<p><strong>Step 1 :</strong> Find $w_{1}$ </p>
<p>Given $z_{1} = 5 + 4i$. </p>
<p>When you rotate $z_{1}$ by $90^{\circ}$ anticlockwise about the origin, the real part becomes negative of the imaginary part of $z_{1}$, and the imaginary part becomes the real part of $z_{1}$.</p>
<p>Therefore, $w_{1}$ becomes :
<br/><br/>$w_{1} = -4 + 5i$</p>
<p><strong>Step 2 :</strong> Find $w_{2}$</p>
<p>Given $z_{2} = 3 + 5i$.</p>
<p>When you rotate $z_{2}$ by $90^{\circ}$ clockwise about the origin, the real part becomes the imaginary part of $z_{2}$, and the imaginary part becomes negative of the real part of $z_{2}$.</p>
<p>Therefore, $w_{2}$ becomes :
<br/><br/>$w_{2} = 5 - 3i$</p>
<p><strong>Step 3 :</strong> Calculate $w_{1} - w_{2}$</p>
<p>$w_{1} - w_{2} = (-4 + 5i) - (5 - 3i)$
<br/><br/>$w_{1} - w_{2} = -9 + 8i$</p>
<p><strong>Step 4 :</strong> Find the principal argument</p>
<p>To find the principal argument, we need to compute the tangent inverse of the ratio of the imaginary part to the real part.
<br/><br/>Argument = $\tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right)$ </p>
<p>Argument = $\tan^{-1}\left(\frac{8}{-9}\right)$
<br/><br/>Argument = $-\tan^{-1}\left(\frac{8}{9}\right)$ </p>
<p>Since it's in the third quadrant, the principal argument is :
<br/><br/>Argument = $\pi + (-\tan^{-1}\left(\frac{8}{9}\right))$
<br/><br/>Argument = $\pi - \tan^{-1}\left(\frac{8}{9}\right)$ </p>
<p>So, the correct option is :
<br/><br/>Option C : $\pi-\tan^{-1} \frac{8}{9}$</p>
|
mcq
|
jee-main-2023-online-11th-april-morning-shift
|
jaoe38c1lse5qzoe
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
<p>If $$\alpha$$ denotes the number of solutions of $$|1-i|^x=2^x$$ and $$\beta=\left(\frac{|z|}{\arg (z)}\right)$$, where $$z=\frac{\pi}{4}(1+i)^4\left[\frac{1-\sqrt{\pi} i}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} i}\right], i=\sqrt{-1}$$, then the distance of the point $$(\alpha, \beta)$$ from the line $$4 x-3 y=7$$ is __________.</p>
|
[]
| null |
3
|
<p>$$\begin{aligned}
& (\sqrt{2})^x=2^x \Rightarrow x=0 \Rightarrow \alpha=1 \\
& z=\frac{\pi}{4}(1+i)^4\left[\frac{\sqrt{\pi}-\pi i-i-\sqrt{\pi}}{\pi+1}+\frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi}}{1+\pi}\right] \\
& =-\frac{\pi i}{2}\left(1+4 i+6 i^2+4 i^3+1\right) \\
& =2 \pi i \\
& \beta=\frac{2 \pi}{\frac{\pi}{2}}=4
\end{aligned}$$</p>
<p>Distance from $$(1,4)$$ to $$4 x-3 y=7$$</p>
<p>Will be $$\frac{15}{5}=3$$</p>
|
integer
|
jee-main-2024-online-31st-january-morning-shift
|
jaoe38c1lsfkvcni
|
maths
|
complex-numbers
|
argument-of-complex-numbers
|
<p>Let $$\mathrm{r}$$ and $$\theta$$ respectively be the modulus and amplitude of the complex number $$z=2-i\left(2 \tan \frac{5 \pi}{8}\right)$$, then $$(\mathrm{r}, \theta)$$ is equal to</p>
|
[{"identifier": "A", "content": "$$\\left(2 \\sec \\frac{11 \\pi}{8}, \\frac{11 \\pi}{8}\\right)$$\n"}, {"identifier": "B", "content": "$$\\left(2 \\sec \\frac{3 \\pi}{8}, \\frac{3 \\pi}{8}\\right)$$\n"}, {"identifier": "C", "content": "$$\\left(2 \\sec \\frac{5 \\pi}{8}, \\frac{3 \\pi}{8}\\right)$$\n"}, {"identifier": "D", "content": "$$\\left(2 \\sec \\frac{3 \\pi}{8}, \\frac{5 \\pi}{8}\\right)$$"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& z=2-i\left(2 \tan \frac{5 \pi}{8}\right)=x+i y(\text { let }) \\
& r=\sqrt{x^2+y^2} ~\& ~\theta=\tan ^{-1} \frac{y}{x} \\
& r=\sqrt{(2)^2+\left(2 \tan \frac{5 \pi}{8}\right)^2} \\
& =\left|2 \sec \frac{5 \pi}{8}\right|=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right| \\
& =2 \sec \frac{3 \pi}{8} \\
& \& ~\theta = {\tan ^{ - 1}}\left( {{{ - 2\tan {{5\pi } \over 8}} \over 2}} \right) \\
& =\tan ^{-1}\left(\tan ^2\left(\pi-\frac{5 \pi}{8}\right)\right) \\
& =\frac{3 \pi}{8}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-29th-january-evening-shift
|
PM4qMlStSHAcQdgu
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
The conjugate of a complex number is $${1 \over {i - 1}}$$ then that complex number is :
|
[{"identifier": "A", "content": "$${{ - 1} \\over {i - 1}}$$ "}, {"identifier": "B", "content": "$${1 \\over {i + 1}}\\,$$ "}, {"identifier": "C", "content": "$${{ - 1} \\over {i + 1}}$$ "}, {"identifier": "D", "content": "$${1 \\over {i - 1}}$$ "}]
|
["C"]
| null |
$$\left( {{1 \over {i - 1}}} \right) = {1 \over { - i - 1}} = {{ - 1} \over {i + 1}}$$
|
mcq
|
aieee-2008
|
I5pK1RCzpUnmvnIyw53rsa0w2w9jwxjodf8
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
If a > 0 and z = $${{{{\left( {1 + i} \right)}^2}} \over {a - i}}$$, has magnitude $$\sqrt {{2 \over 5}} $$, then $$\overline z $$ is equal to :
|
[{"identifier": "A", "content": "$$ - {1 \\over 5} + {3 \\over 5}i$$"}, {"identifier": "B", "content": "$$ - {1 \\over 5} - {3 \\over 5}i$$"}, {"identifier": "C", "content": "$${1 \\over 5} - {3 \\over 5}i$$"}, {"identifier": "D", "content": "$$ - {3 \\over 5} - {1 \\over 5}i$$"}]
|
["B"]
| null |
$$z = {{{{\left( {1 + i} \right)}^2}} \over {a - i}} \times {{a + i} \over {a + i}}$$<br><br>
$$ \Rightarrow z = {{\left( {1 - 1 + 2i} \right)\left( {a + i} \right)} \over {{a^2} + 1}} = {{2ai - 2} \over {{a^2} + 1}}$$ <br><br>
$$ \Rightarrow \left| z \right| = \sqrt {{{\left( {{{ - 2} \over {{a^2} + 1}}} \right)}^2} + {{\left( {{{2a} \over {{a^2} + 1}}} \right)}^2}} = \sqrt {{{4 + 4{a^2}} \over {{{({a^2} + 1)}^2}}}} $$<br><br>
$$ \Rightarrow \sqrt {{{4(1 + {a^2})} \over {{{({a^2} + 1)}^2}}}} = {2 \over {\sqrt {{a^2} + 1} }}$$.... (i)<br><br>
Now given $$\left| z \right| = \sqrt {{2 \over 5}} $$<br><br>
so $$\sqrt {{2 \over 5}} = {2 \over {\sqrt {1 + {a^2}} }}$$ from equation (i)<br><br>
By squaring both sides<br><br>
$$ \Rightarrow {2 \over 5} = {4 \over {1 + {a^2}}}$$<br><br>
$$ \Rightarrow 1 + {a^2} = 10$$<br><br>
a<sup>2</sup> = 9<br><br>
$$ \Rightarrow $$ a = $$ \pm $$ 3
<br><br>$$ \because $$ (a > 0) then a = 3
<br><br>Now $$\overline z = {{{{\left( {1 - i} \right)}^2}} \over {3 + i}}$$
<br><br>= $${{\left( {1 + {i^2} - 2i} \right)} \over {3 + i}}$$
<br><br>= $${{\left( {1 + {i^2} - 2i} \right)\left( {3 - i} \right)} \over {\left( {3 + i} \right)\left( {3 - i} \right)}}$$
<br><br>= $${{\left( { - 2i} \right)\left( {3 - i} \right)} \over {\left( {9 - {i^2}} \right)}}$$
<br><br>= $${{\left( { - 2i} \right)\left( {3 - i} \right)} \over {10}}$$
<br><br>= $${{ - 6i + 2{i^2}} \over {10}}$$
<br><br>= $${{ - 6i - 2} \over {10}}$$
<br><br>= $$-{1 \over 5} - {3 \over 5}i$$
|
mcq
|
jee-main-2019-online-10th-april-morning-slot
|
LNvcjihjOACPdCzciI7k9k2k5grsymq
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
If the equation, x<sup>2</sup> + bx + 45 = 0 (b $$ \in $$ R) has
conjugate complex roots and they satisfy
|z +1| = 2$$\sqrt {10} $$ , then :
|
[{"identifier": "A", "content": "b<sup>2</sup> \u2013 b = 42"}, {"identifier": "B", "content": "b<sup>2</sup> + b = 12"}, {"identifier": "C", "content": "b<sup>2</sup> + b = 72"}, {"identifier": "D", "content": "b<sup>2</sup> \u2013 b = 30"}]
|
["D"]
| null |
x<sup>2</sup>
+ bx = 45 = 0 (b $$ \in $$ R)
<br>has roots $$\alpha $$ + i$$\beta $$, $$\alpha $$ – i$$\beta $$
<br>sum of roots = – b = 2$$\alpha $$
<br>product of roots = 45 =
$$\alpha $$<sup>2</sup>
+ $$\beta $$<sup>2</sup>
<br><br>Let z = x + iy
<br><br>$$ \therefore $$ |x + iy +1| = 2$$\sqrt {10} $$
<br><br>$${\left| {x + iy + 1} \right|^2} = {\left( {2\sqrt {10} } \right)^2}$$
<br><br>$$ \Rightarrow $$ (x + 1)<sup>2</sup> + y<sup>2</sup> = 40
<br><br>$$ \Rightarrow $$ ($$\alpha $$ + 1)<sup>2</sup> + $$\beta $$<sup>2</sup> = 40
<br><br> [putting real part $$\alpha $$ in place of x and imaginary part $$\beta $$ in place of y]
<br><br>$$ \Rightarrow $$ $$\alpha $$<sup>2</sup> + 2$$\alpha $$ + 1 + $$\beta $$<sup>2</sup> = 40
<br><br>$$ \Rightarrow $$ 45 + 2$$\alpha $$ + 1 = 40
<br><br>$$ \Rightarrow $$ $$\alpha $$ = -3
<br><br>$$ \therefore $$ -b = 2$$\alpha $$ = 2$$ \times $$(-3) = -6
<br><br>$$ \Rightarrow $$ b = 6
<br><br>By checking options we found b<sup>2</sup> – b = 30.
|
mcq
|
jee-main-2020-online-8th-january-morning-slot
|
zRF8sEwtoachGxzvEr1klt7y6a1
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
If $$\alpha$$, $$\beta$$ $$\in$$ R are such that 1 $$-$$ 2i (here i<sup>2</sup> = $$-$$1) is a root of z<sup>2</sup> + $$\alpha$$z + $$\beta$$ = 0, then ($$\alpha$$ $$-$$ $$\beta$$) is equal to :
|
[{"identifier": "A", "content": "$$-$$7"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$-$$3"}]
|
["A"]
| null |
1 $$-$$ 2i is the root of the equation. So other root is 1 $$+$$ 2i
<br><br>$$ \therefore $$ Sum of roots = 1 $$-$$ 2i + 1 $$+$$ 2i = 2 = -$$\alpha $$
<br><br>Product of roots = (1 $$-$$ 2i)(1 $$+$$ 2i) = 1 - 4i<sup>2</sup> = 5 = $$\beta $$
<br><br>$$ \therefore $$ $$\alpha $$ - $$\beta $$ = -2 - 5 = -7
|
mcq
|
jee-main-2021-online-25th-february-evening-slot
|
LbJQB0EUlx4IaWnLwY1kmhzri06
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
Let z and $$\omega$$ be two complex numbers such that $$\omega = z\overline z - 2z + 2,\left| {{{z + i} \over {z - 3i}}} \right| = 1$$ and Re($$\omega$$) has minimum value. Then, the minimum value of n $$\in$$ N for which $$\omega$$<sup>n</sup> is real, is equal to ______________.
|
[]
| null |
4
|
Let z = x + iy<br><br>| z + i | = | z $$-$$ 3i |<br><br>$$ \Rightarrow $$ y = 1<br><br>Now <br><br>$$\omega$$ = x<sup>2</sup> + y<sup>2</sup> $$-$$ 2x $$-$$ 2iy + 2<br><br>$$\omega$$ = x<sup>2</sup> + 1 $$-$$ 2x $$-$$ 2i + 2<br><br>Re($$\omega$$) = x<sup>2</sup> $$-$$ 2x + 3<br><br>Re($$\omega$$) = (x $$-$$ 1)<sup>2</sup> + 2<br><br>Re($$\omega$$)<sub>min</sub> at x = 1 $$ \Rightarrow $$ z = 1 + i<br><br>Now, <br><br>$$\omega$$ = 1 + 1 $$-$$ 2 $$-$$ 2i + 2<br><br>$$\omega$$ = 2(1 $$-$$ i) = 2$$\sqrt 2 {e^{i\left( {{{ - \pi } \over 4}} \right)}}$$<br><br>$$\omega$$<sup>n</sup> = 2$$\sqrt 2 {e^{i\left( {{{ - n\pi } \over 4}} \right)}}$$<br><br>If $$\omega$$<sup>n</sup> is real $$ \Rightarrow $$ n = 4
|
integer
|
jee-main-2021-online-16th-march-morning-shift
|
1kru4nljk
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
Let n denote the number of solutions of the equation z<sup>2</sup> + 3$$\overline z $$ = 0, where z is a complex number. Then the value of $$\sum\limits_{k = 0}^\infty {{1 \over {{n^k}}}} $$ is equal to :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${4 \\over 3}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "2"}]
|
["B"]
| null |
z<sup>2</sup> + 3$$\overline z $$ = 0<br><br>Put z = x + iy<br><br>$$\Rightarrow$$ x<sup>2</sup> $$-$$ y<sup>2</sup> + 2ixy + 3(x $$-$$ iy) = 0<br><br>$$\Rightarrow$$ (x<sup>2</sup> $$-$$ y<sup>2</sup> + 3x) + i(2xy $$-$$ 3y) = 0 + i0<br><br>$$\therefore$$ x<sup>2</sup> $$-$$ y<sup>2</sup> + 3x = 0 ..... (1)<br><br>2xy $$-$$ 3y = 0 ..... (2)<br><br>x = $${3 \over 2}$$, y = 0<br><br>Put x = $${3 \over 2}$$ in equation (1)<br><br>$${9 \over 4} - {y^2} + {9 \over 2} = 0$$<br><br>$${y^2} = {{27} \over 4} \Rightarrow y = \pm {{3\sqrt 3 } \over 2}$$<br><br>$$\therefore$$ $$(x,y) = \left( {{3 \over 2},{{3\sqrt 3 } \over 2}} \right),\left( {{3 \over 2},{{ - 3\sqrt 3 } \over 2}} \right)$$<br><br>Put y = 0 $$\Rightarrow$$ x<sup>2</sup> $$-$$ 0 + 3x = 0<br><br>x = 0, $$-$$3<br><br>$$\therefore$$ (x, y) = (0, 0), ($$-$$3, 0)<br><br>$$\therefore$$ No of solutions = n = 4<br><br>$$\sum\limits_{K = 0}^\infty {\left( {{1 \over {{n^k}}}} \right)} = \sum\limits_{K = 0}^\infty {\left( {{1 \over {4{n^k}}}} \right)} $$<br><br>$$ = {1 \over 1} + {1 \over 4} + {1 \over {16}} + {1 \over {64}} + ......$$<br><br>$$ = {1 \over {1 - {1 \over 4}}} = {4 \over 3}$$
|
mcq
|
jee-main-2021-online-22th-july-evening-shift
|
1l55j8k9n
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
<p>Sum of squares of modulus of all the complex numbers z satisfying $$\overline z = i{z^2} + {z^2} - z$$ is equal to ___________.</p>
|
[]
| null |
2
|
Let $z=x+i y$
<br/><br/>
So $2 x=(1+i)\left(x^{2}-y^{2}+2 x y i\right)$
<br/><br/>
$\Rightarrow 2 x=x^{2}-y^{2}-2 x y\quad$ ...(i) and
<br/><br/>
$$
x^{2}-y^{2}+2 x y=0\quad\dots(ii)
$$
<br/><br/>
From (i) and (ii) we get
<br/><br/>
$$
x=0 \text { or } y=-\frac{1}{2}
$$
<br/><br/>
When $x=0$ we get $y=0$
<br/><br/>
When $y=-\frac{1}{2}$ we get $x^{2}-x-\frac{1}{4}=0$
<br/><br/>
$\Rightarrow \quad x=\frac{-1 \pm \sqrt{2}}{2}$
<br/><br/>
So there will be total 3 possible values of $z$, which are $0,\left(\frac{-1+\sqrt{2}}{2}\right)-\frac{1}{2} i$ and $\left(\frac{-1-\sqrt{2}}{2}\right)-\frac{1}{2} i$
<br/><br/>
Sum of squares of modulus
<br/><br/>
$$
\begin{aligned}
&=0+\left(\frac{\sqrt{2}-1}{2}\right)^{2}+\frac{1}{4}+\left(\frac{\sqrt{2}+1}{2}\right)^{2}=+\frac{1}{4} \\\\
&=2
\end{aligned}
$$
|
integer
|
jee-main-2022-online-28th-june-evening-shift
|
1l57ntha9
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
<p>The area of the polygon, whose vertices are the non-real roots of the equation $$\overline z = i{z^2}$$ is :</p>
|
[{"identifier": "A", "content": "$${{3\\sqrt 3 } \\over 4}$$"}, {"identifier": "B", "content": "$${{3\\sqrt 3 } \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "$${3 \\over 4}$$"}]
|
["A"]
| null |
<p>$$\overline z = i{z^2}$$</p>
<p>Let $$z = x + iy$$</p>
<p>$$x - iy = i({x^2} - {y^2} + 2xiy)$$</p>
<p>$$x - iy = i({x^2} - {y^2}) - 2xy$$</p>
<p>$$\therefore$$ $$x = - 2yx$$ or $${x^2} - {y^2} = - y$$</p>
<p>$$x = 0$$ or $$y = - {1 \over 2}$$</p>
<p>Case - I</p>
<p>$$x = 0$$</p>
<p>$$ - {y^2} = - y$$</p>
<p>$$y = 0,\,\,1$$</p>
<p>Case - II</p>
<p>$$y = - {1 \over 2}$$</p>
<p>$$ \Rightarrow {x^2} - {1 \over 4} = {1 \over 2} \Rightarrow x = \pm {{\sqrt 3 } \over 2}$$</p>
<p>$$x = \left\{ {0,i,{{\sqrt 3 } \over 2} - {i \over 2},{{ - \sqrt 3 } \over 2} - {i \over 2}} \right\}$$</p>
<p>Area of polygon $$ = {1 \over 2}\left| {\matrix{
0 & 1 & 1 \cr
{{{\sqrt 3 } \over 2}} & {{{ - 1} \over 2}} & 1 \cr
{{{ - \sqrt 3 } \over 2}} & {{{ - 1} \over 2}} & 1 \cr
} } \right|$$</p>
<p>$$ = {1 \over 2}\left| { - \sqrt 3 \,\,\, - {{\sqrt 3 } \over 2}} \right| = {{3\sqrt 3 } \over 4}$$</p>
|
mcq
|
jee-main-2022-online-27th-june-morning-shift
|
1l5vz2v9s
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
<p>The real part of the complex number $${{{{(1 + 2i)}^8}\,.\,{{(1 - 2i)}^2}} \over {(3 + 2i)\,.\,\overline {(4 - 6i)} }}$$ is equal to :</p>
|
[{"identifier": "A", "content": "$${{500} \\over {13}}$$"}, {"identifier": "B", "content": "$${{110} \\over {13}}$$"}, {"identifier": "C", "content": "$${{55} \\over {6}}$$"}, {"identifier": "D", "content": "$${{550} \\over {13}}$$"}]
|
["D"]
| null |
<p>Given,</p>
<p>$${{{{(1 + 2i)}^8}\,.\,{{(1 - 2i)}^2}} \over {(3 + 2i)\,.\,\overline {(4 - 6i)} }}$$</p>
<p>$$ = {{{{(1 + 2i)}^2}{{(1 - 2i)}^2}{{(1 + 2i)}^6}} \over {(3 + 2i)(4 + 6i)}}$$</p>
<p>$$ = {{{{(1 - 4{i^2})}^2}{{(1 + 2i)}^6}} \over {12 + 18i + 8i + 12{i^2}}}$$</p>
<p>$$ = {{{{(1 + 5)}^2}{{\left[ {{{(1 + 2i)}^2}} \right]}^3}} \over {12 + 26i - 12}}$$</p>
<p>$$ = {{25{{(1 + 4{i^2} + 4i)}^3}} \over {26i}}$$</p>
<p>$$ = {{25{{(1 - 4 + 4i)}^3}} \over {26i}}$$</p>
<p>$$ = {{25{{( - 3 + 4i)}^3}} \over {26i}}$$</p>
<p>$$ = {{25} \over {26i}}\left[ {{{( - 3)}^3} + {{(4i)}^3} + 3\,.\,{{( - 3)}^2}\,.\,4i + 3( - 3)\,.\,{{(4i)}^2}} \right]$$</p>
<p>$$ = {{25} \over {26i}}( - 27 - 64i + 108i + 144)$$</p>
<p>$$ = {{25} \over {26i}}(117 + 44i)$$</p>
<p>$$ = {{25i} \over {26{i^2}}}(117 + 44i)$$</p>
<p>$$ = {{25i} \over { - 26}}(117 + 44i)$$</p>
<p>$$ = {{25 \times 117i} \over { - 26}} - {{25 \times 44{i^2}} \over {26}}$$</p>
<p>$$ = {{25 \times 117i} \over { - 26}} + {{22 \times 25} \over {13}}$$</p>
<p>$$ = {{25 \times 117i} \over { - 26}} + {{550} \over {13}}$$</p>
<p>$$\therefore$$ Real part $$ = {{550} \over {13}}$$</p>
|
mcq
|
jee-main-2022-online-30th-june-morning-shift
|
1l6jennth
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
<p>Let $$S=\left\{z \in \mathbb{C}: z^{2}+\bar{z}=0\right\}$$. Then $$\sum\limits_{z \in S}(\operatorname{Re}(z)+\operatorname{Im}(z))$$ is equal to ______________.</p>
|
[]
| null |
0
|
<p>$$\because$$ $${z^2} + \overline z = 0$$</p>
<p>Let $$z = x + iy$$</p>
<p>$$\therefore$$ $${x^2} - {y^2} + 2ixy + x - iy = 0$$</p>
<p>$$({x^2} - {y^2} + x) + i(2xy - y) = 0$$</p>
<p>$$\therefore$$ $${x^2} + {y^2} = 0$$ and $$(2x - 1)y = 0$$</p>
<p>if $$x = \, + \,{1 \over 2}$$ then $$y = \, \pm \,{{\sqrt 3 } \over 2}$$</p>
<p>And if $$y = 0$$ then $$x = 0, - 1$$</p>
<p>$$\therefore$$ $$z = 0 + 0i, - 1 + 0i,{1 \over 2} + {{\sqrt 3 } \over 2}i,{1 \over 2} - {{\sqrt 3 } \over 2}i$$</p>
<p>$$\therefore$$ $$\sum {\left( {{R_e}(z) + m(z)} \right) = 0} $$</p>
|
integer
|
jee-main-2022-online-27th-july-morning-shift
|
1l6p0k7zx
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
<p>If $$z=2+3 i$$, then $$z^{5}+(\bar{z})^{5}$$ is equal to :</p>
|
[{"identifier": "A", "content": "244"}, {"identifier": "B", "content": "224"}, {"identifier": "C", "content": "245"}, {"identifier": "D", "content": "265"}]
|
["A"]
| null |
<p>$$z = (2 + 3i)$$</p>
<p>$$ \Rightarrow {z^5} = (2 + 3i){\left( {{{(2 + 3i)}^2}} \right)^2}$$</p>
<p>$$ = (2 + 3i){( - 5 + 12i)^2}$$</p>
<p>$$ = (2 + 3i)( - 119 - 120i)$$</p>
<p>$$ = - 238 - 240i - 357i + 360$$</p>
<p>$$ = 122 - 597i$$</p>
<p>$${\overline z ^5} = 122 + 597i$$</p>
<p>$${z^5} + {\overline z ^5} = 244$$</p>
|
mcq
|
jee-main-2022-online-29th-july-morning-shift
|
lgnyjdfm
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
If the set $\left\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in \mathbb{C}, \operatorname{Re}(z)=3\right\}$ is equal to<br/><br/> the interval $(\alpha, \beta]$, then $24(\beta-\alpha)$ is equal to :
|
[{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "27"}, {"identifier": "C", "content": "42"}, {"identifier": "D", "content": "30"}]
|
["D"]
| null |
Let $z_1=\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right)$<br/><br/>
Let $\mathrm{z}=3+\mathrm{iy}$<br/><br/> $\bar{z}=3-i y$<br/><br/>
$$
\begin{aligned}
& z_1=\frac{2 i y+\left(9+y^2\right)}{2-3(3+i y)+5(3-i y)} \\\\
& =\frac{9+y^2+i(2 y)}{8-8 i y} \\\\
& =\frac{\left(9+y^2\right)+i(2 y)}{8(1-i y)} \\\\
& \operatorname{Re}\left(z_1\right)=\frac{\left(9+y^2\right)-2 y^2}{8\left(1+y^2\right)}
\end{aligned}
$$<br/><br/>
$$
\begin{aligned}
& =\frac{9-y^2}{8\left(1+y^2\right)} \\\\
& =\frac{1}{8}\left[\frac{10-\left(1+y^2\right)}{\left(1+y^2\right)}\right] \\\\
& =\frac{1}{8}\left[\frac{10}{1+y^2}-1\right] \\\\
& 1+y^2 \in[1, \infty] \\\\
& \frac{1}{1+y^2} \in(0,1] \\\\
& \frac{10}{1+y^2} \in(0,10] \\\\
& \frac{10}{1+y^2}-1 \in(-1,9] \\\\
& \operatorname{Re}\left(\mathrm{z}_1\right) \in\left(\frac{-1}{8}, \frac{9}{8}\right] \\\\
& \alpha=\frac{-1}{8}, \beta=\frac{9}{8} \\\\
& 24(\beta-\alpha)=24\left(\frac{9}{8}+\frac{1}{8}\right)=30
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-15th-april-morning-shift
|
1lgow71ev
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
<p>Let $$S=\left\{z \in \mathbb{C}: \bar{z}=i\left(z^{2}+\operatorname{Re}(\bar{z})\right)\right\}$$. Then $$\sum_\limits{z \in \mathrm{S}}|z|^{2}$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\frac{7}{2}$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$\\frac{5}{2}$$"}]
|
["B"]
| null |
Let $z=x+i y$
<br/><br/>$$
\begin{aligned}
&\bar{z}=i\left(z^2+\operatorname{Re}(\bar{z})\right) \\\\
&\Rightarrow x-i y=i\left(x^2-y^2+2 i x y+x\right) \\\\
& x-i y=i\left(x^2-y^2+x\right)-2 x y \\\\
&x=-2 x y \Rightarrow x(2 y+1)=0 \\\\
&\Rightarrow x=0, y=\frac{-1}{2} ........(i)\\\\
&-y=x^2-y^2+x ...........(ii)
\end{aligned}
$$
<br/><br/><b>Case (I)</b> $x=0$
<br/><br/>$\Rightarrow-y=-y^2 \Rightarrow y^2-y=0 \Rightarrow y=0,1$
<br/><br/>$$ \therefore $$ $z=0, i$
<br/><br/><b>Case (II)</b> $y=\frac{-1}{2}$
<br/><br/>$\Rightarrow \frac{1}{2}=x^2-\frac{1}{4}+x \Rightarrow x^2+x-\frac{3}{4}=0$
<br/><br/>$$ \Rightarrow $$ $4 x+4 x-3=0 \Rightarrow(2 x-1)(2 x+3)=0$
<br/><br/>$$ \Rightarrow $$ $x=\frac{1}{2}, \frac{-3}{2}$
<br/><br/>$$
\begin{aligned}
& z=\frac{1}{2}-\frac{1}{2} i, \frac{-3}{2}-\frac{1}{2} i \\\\
& \sum|z|^2=0+1+\frac{1}{2}+\frac{5}{2}=4
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-13th-april-evening-shift
|
1lgsugbnn
|
maths
|
complex-numbers
|
conjugate-of-complex-numbers
|
<p>For $$a \in \mathbb{C}$$, let $$\mathrm{A}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z}) > \operatorname{Im}(\bar{a}+z)\}$$ and $$\mathrm{B}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)\}$$. Then among the two statements :</p>
<p>(S1): If $$\operatorname{Re}(a), \operatorname{Im}(a) > 0$$, then the set A contains all the real numbers</p><p>(S2) : If $$\operatorname{Re}(a), \operatorname{Im}(a) < 0$$, then the set B contains all the real numbers,</p>
|
[{"identifier": "A", "content": "both are false"}, {"identifier": "B", "content": "only (S1) is true"}, {"identifier": "C", "content": "only (S2) is true"}, {"identifier": "D", "content": "both are true"}]
|
["A"]
| null |
We are given that $a \in \mathbb{C}$ and $z \in \mathbb{C}$.
<br/><br/>Let $a = x_1 + iy_1$ and $z = x_2 + iy_2$ where $x_1, y_1, x_2, y_2 \in \mathbb{R}$
<br/><br/>We are also given two sets A and B defined as follows :
<br/><br/>- A is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is greater than the imaginary part of $(\overline{a} + z)$.
<br/><br/>- B is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is less than the imaginary part of $(\overline{a} + z)$.
<br/><br/>Statement (S1) says : If the real part and imaginary part of $a$ are both positive, then the set A contains all the real numbers.
<br/><br/>Statement (S2) says : If the real part and imaginary part of $a$ are both negative, then the set B contains all the real numbers.
<br/><br/>We need to determine which of these statements are true.
<br/><br/>Let's evaluate each statement.
<br/><br/>1. Statement (S1) : For $z \in A$,
<br/><br/> $Re(a + \overline{z}) > Im(\overline{a} + z)$
<br/><br/> This can be re-written as $x_1 + x_2 > y_2 - y_1$
<br/><br/> If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 > 0$, then the condition simplifies to $x_2 > -(x_1 + y_1)$.
<br/><br/> This indicates that A covers a part of the negative real axis, but not the entire real axis. Therefore, Statement (S1) is false.
<br/><br/>2. Statement (S2) : For $z \in B$,
<br/><br/> $Re(a + \overline{z}) < Im(\overline{a} + z)$
<br/><br/> This can be re-written as $x_1 + x_2 < y_2 - y_1$
<br/><br/> If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 < 0$, then the condition simplifies to $x_2 < -(x_1 + y_1)$.
<br/><br/> This indicates that B covers a part of the positive real axis, but not the entire real axis. Therefore, Statement (S2) is false.
<br/><br/>Therefore, both (S1) and (S2) are false, so the answer is Option A : both are false.
|
mcq
|
jee-main-2023-online-11th-april-evening-shift
|
rPNBPY8xjQoyu1Iq
|
maths
|
complex-numbers
|
cube-roots-of-unity
|
If $$z = x - iy$$ and $${z^{{1 \over 3}}} = p + iq$$, then
<br/><br/>$${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$ is equal to :
|
[{"identifier": "A", "content": "- 2"}, {"identifier": "B", "content": "- 1"}, {"identifier": "C", "content": "2 "}, {"identifier": "D", "content": "1"}]
|
["A"]
| null |
Given $${z^{{1 \over 3}}} = p + iq$$
<br><br>$$ \Rightarrow $$ z = (p + iq)<sup>3</sup>
<br><br>= p<sup>3</sup> + (iq)<sup>3</sup> +3p(iq)(p + iq)
<br><br>= p<sup>3</sup> - iq<sup>3</sup> +3ip<sup>2</sup>q - 3pq<sup>2</sup>
<br><br>= p(p<sup>2</sup> - 3q<sup>2</sup>) - iq(q<sup>2</sup> - 3p<sup>2</sup>)
<br><br>Given that $$z = x - iy$$
<br><br>$$\therefore$$ $$x - iy$$ = p(p<sup>2</sup> - 3q<sup>2</sup>) - iq(q<sup>2</sup> - 3p<sup>2</sup>)
<br><br>By comparing both sides we get,
<br><br>$${x \over p} = {p^2} - 3{q^2}$$ and $${y \over q} = {q^2} - 3{p^2}$$
<br><br>$$\therefore$$ $${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$
<br><br>= $${{{p^2} - 3{q^2} + {q^2} - 3{p^2}} \over {{p^2} + {q^2}}}$$
<br><br>= $${{ - 2{q^2} - 2{p^2}} \over {{p^2} + {q^2}}}$$
<br><br>= $${{ - 2\left( {{q^2} + {p^2}} \right)} \over {{p^2} + {q^2}}}$$
<br><br>= $$-2$$
|
mcq
|
aieee-2004
|
b3Gq6JEgA0LBu57T
|
maths
|
complex-numbers
|
cube-roots-of-unity
|
If the cube roots of unity are 1, $$\omega \,,\,{\omega ^2}$$ then the roots of the equation $${(x - 1)^3}$$ + 8 = 0, are :
|
[{"identifier": "A", "content": "$$ - 1, - 1 + 2\\,\\,\\omega , - 1 - 2\\,\\,{\\omega ^2}$$ "}, {"identifier": "B", "content": "$$ - 1, - 1, - 1$$ "}, {"identifier": "C", "content": "$$ - 1,1 - 2\\omega ,1 - 2{\\omega ^2}$$ "}, {"identifier": "D", "content": "$$ - 1,1 + 2\\omega ,1 + 2{\\omega ^2}$$ "}]
|
["C"]
| null |
$${\left( {x - 1} \right)^3} + 8 = 0$$
<br><br>$$ \Rightarrow \left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{1/3}}$$
<br><br>$$ \Rightarrow x - 1 = - 2\,\,\,$$ or $$\,\,\, - 2\omega \,\,\,\,$$ or $$\,\,\,\, - 2{\omega ^2}$$
<br><br>or $$\,\,\,x = - 1\,\,\,$$ or $$\,\,\,1 - 2\omega \,\,\,$$ or $$\,\,\,1 - 2{\omega ^2}.$$
|
mcq
|
aieee-2005
|
4T8jQpKN8qkvEKvH
|
maths
|
complex-numbers
|
cube-roots-of-unity
|
If $$\omega ( \ne 1)$$ is a cube root of unity, and $${(1 + \omega )^7} = A + B\omega \,$$. Then $$(A,B)$$ equals :
|
[{"identifier": "A", "content": "(1 ,1)"}, {"identifier": "B", "content": "(1, 0)"}, {"identifier": "C", "content": "(- 1 ,1)"}, {"identifier": "D", "content": "(0 ,1)"}]
|
["A"]
| null |
$${\left( {1 + \omega } \right)^7} = A + B\omega ;\,\,\,\,{\left( { - {\omega ^2}} \right)^7} = A + B\omega $$
<br><br>$$ - {\omega ^2} = A + B\omega ;\,\,\,\,\,\,\,\,\,\,1 + \omega = A + B\omega $$
<br><br>$$ \Rightarrow A = 1,B = 1.$$
|
mcq
|
aieee-2011
|
15Ymvfe9vqnI6Ico
|
maths
|
complex-numbers
|
cube-roots-of-unity
|
If $$\alpha ,\beta \in C$$ are the distinct roots of the equation
<br/>x<sup>2</sup> - x + 1 = 0, then $${\alpha ^{101}} + {\beta ^{107}}$$ is equal to :
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "-1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}]
|
["D"]
| null |
Given equation,
<br><br>x<sup>2</sup> $$-$$ x + 1 = 0
<br><br>Roots of this equation
<br><br>x = $${{1 \pm \sqrt 3 i} \over 2}$$
<br><br>$$\therefore\,\,\,$$ $$ \propto \, = \,{{1 + \sqrt 3 \,i} \over 2}$$
<br><br>and $$\beta = \,{{1 - \sqrt 3 \,i} \over 2}$$
<br><br>We know;
<br><br>$$\omega = {{ - 1 + \sqrt 3 \,i} \over 2} = - \left( {{{1 - \sqrt 3 \,i} \over 2}} \right) = - \beta $$
<br><br>and $${\omega ^2} = {{ - 1 - \sqrt 3 \,i} \over 2} = - \left( {{{1 + \sqrt 3 \,i} \over 2}} \right) = - \propto $$
<br><br>$$\therefore\,\,\,$$ $$ \propto \, = - {\omega ^2}$$ and $$\beta \, = \, - \omega $$
<br><br>$$\therefore\,\,\,$$ $${ \propto ^{101}} + {\beta ^{107}}$$
<br><br>$$ = {\left( { - {\omega ^2}} \right)^{101}} + {\left( { - \omega } \right)^{107}}$$
<br><br>$$ = {\left( { - 1} \right)^{101}}.{\left( {{\omega ^2}} \right)^{101}} + {\left( { - 1} \right)^{107}}.{\left( \omega \right)^{107}}$$
<br><br>$$ = - 1.{\left( {{\omega ^2}} \right)^{101}} - {\omega ^{107}}$$
<br><br>$$ = - \left( {{\omega ^{202}} + {\omega ^{107}}} \right)$$
<br><br>$$ = - \left( {{\omega ^{3.67}}.\omega + {\omega ^{3.35}}.{\omega ^2}} \right)$$
<br><br>$$ = - \left( {\omega + {\omega ^2}} \right)\,\,\,$$ [ as $$\,\,\,$$ $${\omega ^{3n}} = 1$$]
<br><br>$$ = - \left( { - 1} \right)$$ $$\,\,\,\,\,\,$$ [as $$\,\,\,$$ $$1 + \omega + {\omega ^2} = 0$$ ]
<br><br>$$ = 1$$
|
mcq
|
jee-main-2018-offline
|
5MdWZd7KqONKCwZmNrjgy2xukfagwe44
|
maths
|
complex-numbers
|
cube-roots-of-unity
|
If a and b are real numbers such that<br/> $${\left( {2 + \alpha } \right)^4} = a + b\alpha $$ <br/>where $$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$ then a + b is
equal to :
|
[{"identifier": "A", "content": "33"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "57"}]
|
["B"]
| null |
$$\alpha = \omega $$ as given $$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$
<br><br>$$ \Rightarrow {(2 + \omega )^4} = a + b\omega \,({\omega ^3} = 1)$$<br><br>$$ \Rightarrow {2^4} + {4.2^3}\omega + {6.2^2}{\omega ^3} + 4.2.\,{\omega ^3} + {\omega ^4} = a + b\omega $$<br><br>$$ \Rightarrow 16 + 32\omega + 24{\omega ^2} + 8 + \omega = a + b\omega $$<br><br>$$ \Rightarrow 24 + 24{\omega ^2} + 33\omega = a + b\omega $$<br><br><br>$$ \Rightarrow - 24\omega + 33\omega = a + b\omega $$<br><br>$$ \Rightarrow a = 0,\,b = 9$$
|
mcq
|
jee-main-2020-online-4th-september-evening-slot
|
ZDaEWjLQBunSbbEG
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
If $${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$ then :
|
[{"identifier": "A", "content": "x = 2n + 1, where n is any positive integer"}, {"identifier": "B", "content": "x = 4n , where n is any positive integer"}, {"identifier": "C", "content": "x = 2n, where n is any positive integer\n"}, {"identifier": "D", "content": "x = 4n + 1, where n is any positive integer.\n"}]
|
["B"]
| null |
$${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$
<br><br>$$ \Rightarrow $$ $${\left[ {{{\left( {1 + i} \right)\left( {1 + i} \right)} \over {\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^x} = 1$$
<br><br>$$ \Rightarrow $$ $${\left[ {{{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}}} \right]^x} = 1$$
<br><br>$$ \Rightarrow $$ $${\left[ {{{1 + 2i + {i^2}} \over {1 + 1}}} \right]^x} = 1$$
<br><br>$$ \Rightarrow $$ $${\left[ {{{1 + 2i - 1} \over 2}} \right]^x} = 1$$
<br><br>$$ \Rightarrow {\left( i \right)^x} = 1$$
<br><br>We know $$i = \sqrt { - 1} $$
<br><br>$$\therefore$$ $${i^2} = - 1$$
<br><br>$$ \Rightarrow $$ $${i^3} = - 1 \times i = - i$$
<br><br>$$ \Rightarrow $$ $${i^4} = - i \times i = - {i^2} = - \left( { - 1} \right) = 1$$
<br><br>So when power of $$i$$ is 4 or multiple of 4 then it's value is = 1
<br><br>$$\therefore$$ $${\left( i \right)^x} = 1$$ $$ = {\left( i \right)^{4n}}$$ where n is a positive integer.
|
mcq
|
aieee-2003
|
iKRTnQicxSDJmfsnrrpYO
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
The least positive integer n for which $${\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1,$$ is :
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}]
|
["B"]
| null |
$${\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1$$
<br><br>$$ \Rightarrow \,\,\,\,\,{\left( {{{{{1 + i\sqrt 3 } \over 2}} \over {{{1 - i\sqrt 3 } \over 2}}}} \right)^n} = 1$$
<br><br>We know, $$\omega $$ = $$-$$ $${{1 - i\sqrt 3 } \over 2}$$
<br><br>and $$\omega $$<sup>2</sup> = $$-$$ $${{\left( {1 + i\sqrt 3 } \right)} \over 2}$$
<br><br>$$ \Rightarrow $$ $$\,\,\,\,$$ $${\left( {{{ - {\omega ^2}} \over { - \omega }}} \right)^n} = 1$$
<br><br>$$ \Rightarrow $$ ($$\omega $$)<sup>n</sup> = 1 = $$\omega $$<sup>3</sup>
<br><br>$$ \Rightarrow $$ $$\,\,\,\,$$ n = 3
|
mcq
|
jee-main-2018-online-16th-april-morning-slot
|
xAltR1yM4WVfFWDJ7KLTZ
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
Let z<sub>0</sub> be a root of the quadratic equation, x<sup>2</sup> + x + 1 = 0, If z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$, then arg z is equal to :
|
[{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 6}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "0"}]
|
["A"]
| null |
1 + x + x<sup>2</sup> = 0
<br><br>x = $${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$$
<br><br>z<sub>0</sub> = w, w<sup>2</sup>
<br><br>Now
<br><br>z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$
<br><br>z = 3 + 6iw<sup>81</sup> $$-$$ 3iw<sup>93</sup> (w<sup>93</sup> = w<sup>81</sup> = 1)
<br><br>$$ \Rightarrow $$ z = 3 + 3i
<br><br>then arg(z) = tan<sup>$$-$$1</sup>$$\left( {{3 \over 3}} \right)$$ = tan<sup>$$-$$1</sup> (1) = $${\pi \over 4}$$
|
mcq
|
jee-main-2019-online-9th-january-evening-slot
|
241XVRmztPvAOIbOcO5fL
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
Let $$z = {\left( {{{\sqrt 3 } \over 2} + {i \over 2}} \right)^5} + {\left( {{{\sqrt 3 } \over 2} - {i \over 2}} \right)^5}.$$ If R(z) and 1(z) respectively denote the real and imaginary parts of z, then :
|
[{"identifier": "A", "content": "R(z) = $$-$$ \n3"}, {"identifier": "B", "content": "R(z) < 0 and I(z) > 0"}, {"identifier": "C", "content": "I(z) = 0"}, {"identifier": "D", "content": "R(z) > 0 and I(z) > 0"}]
|
["C"]
| null |
$$z = {\left( {{{\sqrt 3 + i} \over 2}} \right)^5} + {\left( {{{\sqrt 3 - i} \over 2}} \right)^5}$$
<br><br>$$z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}$$
<br><br>$$ = {e^{i5\pi /6}} + {e^{ - i5\pi /6}}$$
<br><br>$$ = \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {{{ - 5\pi } \over 6}} \right) + i\sin \left( {{{ - 5\pi } \over 6}} \right)$$
<br><br>$$ = 2\cos {{5\pi } \over 6} < 0$$
<br><br>$${\rm I}(z) = 0$$ and $${\mathop{\rm Re}\nolimits} (z) < 0$$
|
mcq
|
jee-main-2019-online-10th-january-evening-slot
|
GpTWf3x6bsc54rlvWQCvy
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
Let $${\left( { - 2 - {1 \over 3}i} \right)^3} = {{x + iy} \over {27}}\left( {i = \sqrt { - 1} } \right),\,\,$$ where x and y are real numbers, then y $$-$$ x equals :
|
[{"identifier": "A", "content": "$$-$$ 85 "}, {"identifier": "B", "content": "85"}, {"identifier": "C", "content": "$$-$$ 91 "}, {"identifier": "D", "content": "91"}]
|
["D"]
| null |
$${\left( { - 2 - {i \over 3}} \right)^3} = - {{{{\left( {6 + i} \right)}^3}} \over {27}}$$
<br><br>$$ = {{ - 198 - 107i} \over {27}} = {{x + iy} \over {27}}$$
<br><br>Hence, $$y - x = 198 - 107 = 91$$
|
mcq
|
jee-main-2019-online-11th-january-morning-slot
|
3MeWrbrvU6nSZ6rQ0zoop
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
If $$z = {{\sqrt 3 } \over 2} + {i \over 2}\left( {i = \sqrt { - 1} } \right)$$,
<br/><br/>then (1 + iz + z<sup>5</sup> + iz<sup>8</sup>)<sup>9</sup> is equal to :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "\u20131"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "(-1 + 2i)<sup>9</sup>"}]
|
["B"]
| null |
$$z = {{\sqrt 3 } \over 2} + {i \over 2}$$
<br><br>$$ \Rightarrow $$ z = $$\cos {\pi \over 6}$$ + i $$\sin {\pi \over 6}$$
<br><br>$$ \Rightarrow $$ z = $${e^{i{\pi \over 6}}}$$
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267414/exam_images/mjophfx1aur2l772kqoy.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264578/exam_images/f3bk2z35aewbatcl5s0q.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266661/exam_images/dimoulbitohidywxlo1c.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267369/exam_images/eaqlzkd3lntv18dvvsnh.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264988/exam_images/efub0lbcxf1wsva7rk5z.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264788/exam_images/mnwzigqv4ghhfrkgnyc5.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266440/exam_images/jmhpikhzixbjcrj5tn9t.webp"><source media="(max-width: 1580px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265043/exam_images/qjc2etkouztuvpzl7qbl.webp"><source media="(max-width: 1760px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266331/exam_images/jyagfdzfjyzlc4xbndrp.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266283/exam_images/hlzu8arjd1x3xqvwwhpr.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Evening Slot Mathematics - Complex Numbers Question 125 English Explanation"></picture>
|
mcq
|
jee-main-2019-online-8th-april-evening-slot
|
LLy49LlJXzSsk8gnfijgy2xukewspm3b
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
The value of <br/><br/>$${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$$ is :
|
[{"identifier": "A", "content": "$${1 \\over 2}\\left( {\\sqrt 3 - i} \\right)$$"}, {"identifier": "B", "content": "-$${1 \\over 2}\\left( {\\sqrt 3 - i} \\right)$$"}, {"identifier": "C", "content": "$$ - {1 \\over 2}\\left( {1 - i\\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {1 - i\\sqrt 3 } \\right)$$"}]
|
["B"]
| null |
$${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$$
<br><br>= $${\left( {{{1 + \sin \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right) + i\cos \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right)} \over {1 + \sin \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right) - i\cos \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right)}}} \right)^3}$$
<br><br>= $${\left( {{{1 + \cos \left( {{{5\pi } \over {18}}} \right) + i\sin \left( {{{5\pi } \over {18}}} \right)} \over {1 + \cos \left( {{{5\pi } \over {18}}} \right) - i\sin \left( {{{5\pi } \over {18}}} \right)}}} \right)^3}$$
<br><br>= $${\left( {{{2{{\cos }^2}\left( {{{5\pi } \over {36}}} \right) + 2i\sin \left( {{{5\pi } \over {36}}} \right)\cos \left( {{{5\pi } \over {36}}} \right)} \over {2{{\cos }^2}\left( {{{5\pi } \over {36}}} \right) - 2i\sin \left( {{{5\pi } \over {36}}} \right)\cos \left( {{{5\pi } \over {36}}} \right)}}} \right)^3}$$
<br><br>= $${\left( {{{\cos \left( {{{5\pi } \over {36}}} \right) + i\sin \left( {{{5\pi } \over {36}}} \right)} \over {\cos \left( {{{5\pi } \over {36}}} \right) - i\sin \left( {{{5\pi } \over {36}}} \right)}}} \right)^3}$$
<br><br>= $${\left( {{{{e^{i{{5\pi } \over {36}}}}} \over {{e^{ - i{{5\pi } \over {36}}}}}}} \right)^3}$$
<br><br>= $${\left( {{e^{i{{5\pi } \over {18}}}}} \right)^3}$$
<br><br>= $${{e^{i{{5\pi } \over {18}} \times 3}}}$$ = $${{e^{i{{5\pi } \over 6}}}}$$
<br><br>= $$\cos {{5\pi } \over 6} + i\sin {{5\pi } \over 6}$$
<br><br>= $$ - {{\sqrt 3 } \over 2} + {i \over 2}$$
|
mcq
|
jee-main-2020-online-2nd-september-morning-slot
|
jZ7ETjEkx74JNPhvsujgy2xukfqax1w7
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
The value of $${\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}}$$ is :
|
[{"identifier": "A", "content": "\u20132<sup>15</sup>i"}, {"identifier": "B", "content": "\u20132<sup>15</sup>"}, {"identifier": "C", "content": "2<sup>15</sup>i"}, {"identifier": "D", "content": "6<sup>5</sup>"}]
|
["A"]
| null |
$${\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}}$$
<br><br>= $${\left( {{{2\omega } \over {1 - i}}} \right)^{30}}$$
<br><br>= $${{{2^{30}}.{\omega ^{30}}} \over {{{\left( {{{\left( {1 - i} \right)}^2}} \right)}^{15}}}}$$
<br><br>= $${{{2^{30}}.1} \over {{{\left( {1 + {i^{^2}} - 2i} \right)}^{15}}}}$$
<br><br>= $${{{2^{30}}.1} \over { - {2^{15}}.{i^{15}}}}$$
<br><br>= –2<sup>15</sup>i
|
mcq
|
jee-main-2020-online-5th-september-evening-slot
|
ZH8p2PoFt0hgYa1xTs1klrms6yw
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
Let $$i = \sqrt { - 1} $$. If $${{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{(1 - i)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{(1 + i)}^{24}}}} = k$$, and $$n = [|k|]$$ be the greatest integral part of | k |. Then $$\sum\limits_{j = 0}^{n + 5} {{{(j + 5)}^2} - \sum\limits_{j = 0}^{n + 5} {(j + 5)} } $$ is equal to _________.
|
[]
| null |
310
|
$${(1 + i)^2} = 1 + {i^2} + 2i = 1 - 1 + 2i = 2i$$<br><br>$${(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i$$<br><br>We know,<br><br>$$ - {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega $$<br><br>$$ \Rightarrow - 1 + i\sqrt 3 = 2\omega $$<br><br>and $$ - {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2}$$<br><br>$$ \Rightarrow - 1 - i\sqrt 3 = 2{\omega ^2}$$<br><br>$$ \Rightarrow 1 + i\sqrt 3 = - 2{\omega ^2}$$<br><br>Now, $$K = {{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 - i} \right)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 + i} \right)}^{24}}}}$$<br><br>$$ = {{{{(2\omega )}^{21}}} \over {{{\left( {{{(1 - i)}^2}} \right)}^{12}}}} + {{{{( - 2\omega )}^{21}}} \over {{{\left( {{{(1 + i)}^2}} \right)}^{12}}}}$$<br><br>$$ = {{{2^{21}}.{\omega ^{21}}} \over {{{( - 2i)}^{12}}}} + {{{{( - 2)}^{21}}{{({\omega ^2})}^{21}}} \over {{{(2i)}^{12}}}}$$ [as $${\omega ^3} = 1$$, $${i^4} = 1$$]<br><br>$$ = {{{2^{21}}} \over {{2^{12}}}} - {{{2^{21}}} \over {{2^{12}}}} = 0$$<br><br>$$ \therefore $$ $$n = \left[ {|K|} \right] = \left[ {|0|} \right] = 0$$<br><br>Now $$\sum\limits_{j = 0}^5 {{{(j + 5)}^2}} - \sum\limits_{j = 0}^5 {(j + 5)} $$<br><br>= $$\sum\limits_{j = 0}^5 {({j^2} + 25 + 10j - j - 5)} $$<br><br>= $$\sum\limits_{j = 0}^5 {({j^2} + 9j + 20)} $$<br><br>= $$\sum\limits_{j = 0}^5 {{j^2}} + 9\sum\limits_{j = 0}^5 {j + 20\sum\limits_{j = 0}^5 1 } $$<br><br>= $${{5 \times 6 \times 11} \over 6} + 9\left( {{{5 \times 6} \over 2}} \right) + 20 \times 6$$<br><br>= 55 + 135 + 120<br><br>= 310
|
integer
|
jee-main-2021-online-24th-february-evening-slot
|
1ktd1lmo6
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
If $${\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq)$$, then p and q are roots of the equation :
|
[{"identifier": "A", "content": "$${x^2} - \\left( {\\sqrt 3 - 1} \\right)x - \\sqrt 3 = 0$$"}, {"identifier": "B", "content": "$${x^2} + \\left( {\\sqrt 3 + 1} \\right)x + \\sqrt 3 = 0$$"}, {"identifier": "C", "content": "$${x^2} + \\left( {\\sqrt 3 - 1} \\right)x - \\sqrt 3 = 0$$"}, {"identifier": "D", "content": "$${x^2} - \\left( {\\sqrt 3 + 1} \\right)x + \\sqrt 3 = 0$$"}]
|
["A"]
| null |
$${\left( {2{e^{i\pi /6}}} \right)^{100}} = {2^{99}}(p + iq)$$<br><br>$${2^{100}}\left( {\cos {{50\pi } \over 3} + i\sin {{50\pi } \over 3}} \right) = {2^{99}}(p + iq)$$<br><br>$$p + iq = 2\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)$$<br><br>p = $$-$$1, q = $$\sqrt 3 $$<br><br>$${x^2} - (\sqrt 3 - 1)x - \sqrt 3 = 0$$
|
mcq
|
jee-main-2021-online-26th-august-evening-shift
|
1ktd50jva
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
The least positive integer n such that $${{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}},i = \sqrt { - 1} $$ is a positive integer, is ___________.
|
[]
| null |
6
|
$${{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}} = {{{{(2i)}^n}} \over {{{( - 2i)}^{{{n - 2} \over 2}}}}}$$<br><br>$$ = {{{{(2i)}^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}} = {{{2^{{{n + 2} \over 2}}};{i^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}}$$<br><br>This is positive integer for n = 6
|
integer
|
jee-main-2021-online-26th-august-evening-shift
|
ldo8iriz
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
The complex number $z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ is equal to :
|
[{"identifier": "A", "content": "$\\cos \\frac{\\pi}{12}-i \\sin \\frac{\\pi}{12}$"}, {"identifier": "B", "content": "$\\sqrt{2}\\left(\\cos \\frac{\\pi}{12}+i \\sin \\frac{\\pi}{12}\\right)$"}, {"identifier": "C", "content": "$\\sqrt{2} i\\left(\\cos \\frac{5 \\pi}{12}-i \\sin \\frac{5 \\pi}{12}\\right)$"}, {"identifier": "D", "content": "$\\sqrt{2}\\left(\\cos \\frac{5 \\pi}{12}+i \\sin \\frac{5 \\pi}{12}\\right)$"}]
|
["D"]
| null |
$\mathrm{Z}=\frac{\mathrm{i}-1}{\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}}=\frac{\mathrm{i}-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}}$
<br/><br/>$=\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}} \times \frac{\frac{1}{2}-\sqrt{\frac{3}{2}} \mathrm{i}}{\frac{1}{2}-\sqrt{3 / 2} \mathrm{i}}=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2} \mathrm{i}$
<br/><br/>Apply polar form,
<br/><br/>$r \cos \theta=\frac{\sqrt{3}-1}{2}$
<br/><br/>$r \sin \theta=\frac{\sqrt{3}+1}{2}$
<br/><br/>Now, $\tan \theta=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
<br/><br/>So, $ \theta=\frac{5 \pi}{12}$
|
mcq
|
jee-main-2023-online-31st-january-evening-shift
|
1ldwx26gd
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
<p>The value of $${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$$ is</p>
|
[{"identifier": "A", "content": "$$ - {1 \\over 2}\\left( {1 - i\\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$ - {1 \\over 2}\\left( {\\sqrt 3 - i} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over 2}\\left( {1 - i\\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {\\sqrt 3 + i} \\right)$$"}]
|
["B"]
| null |
$z=\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^{3}$
<br/><br/>
$$
\begin{aligned}
& 1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}=1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18} \\\\
& =1+2 \cos ^{2} \frac{5 \pi}{36}-1+2 i \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36} \\\\
& =2 \cos \frac{5 \pi}{36}\left(\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}\right)=2 \cos \frac{5 \pi}{36} e^{i \frac{5 \pi}{36}} \\\\
& z=-\frac{\sqrt{3}}{2}+\frac{1}{2} i=\frac{1}{2}(i-\sqrt{3})=-\frac{1}{2}(\sqrt{3}-i)
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-24th-january-evening-shift
|
1ldyaoer4
|
maths
|
complex-numbers
|
de-moivre's-theorem
|
<p>Let $$\mathrm{p,q\in\mathbb{R}}$$ and $${\left( {1 - \sqrt 3 i} \right)^{200}} = {2^{199}}(p + iq),i = \sqrt { - 1} $$ then $$\mathrm{p+q+q^2}$$ and $$\mathrm{p-q+q^2}$$ are roots of the equation.</p>
|
[{"identifier": "A", "content": "$${x^2} + 4x - 1 = 0$$"}, {"identifier": "B", "content": "$${x^2} - 4x + 1 = 0$$"}, {"identifier": "C", "content": "$${x^2} + 4x + 1 = 0$$"}, {"identifier": "D", "content": "$${x^2} - 4x - 1 = 0$$"}]
|
["B"]
| null |
<p>$${\left( {1 - \sqrt 3 i} \right)^{200}}$$</p>
<p>$$ = {\left[ {2\left( {{1 \over 2} - {{\sqrt 3 } \over 2}i} \right)} \right]^{200}}$$</p>
<p>$$ = {2^{200}}{\left( {\cos {\pi \over 3} - i\sin {\pi \over 3}} \right)^{200}}$$</p>
<p>$$ = {2^{200}}\left( {\cos {{200\pi } \over 3} - i\sin {{200\pi } \over 3}} \right)$$</p>
<p>$$ = {2^{200}}\left( {\cos \left( {66\pi + {{2\pi } \over 3}} \right) - i\sin \left( {66\theta + {{2\pi } \over 3}} \right)} \right)$$</p>
<p>$$ = {2^{200}}\left( {\cos {{2\pi } \over 3} - i\sin {{2\pi } \over 3}} \right)$$</p>
<p>$$ = {2^{200}}\left( { - {1 \over 2} - {{\sqrt 3 } \over 2}i} \right)$$</p>
<p>$$ = {2^{199}}\left( { - 1 - \sqrt 3 i} \right)$$</p>
<p>$$ = {2^{199}}\left( {p + iq} \right)$$</p>
<p>$$\therefore$$ p = $$-$$1 and q = $$- \sqrt3$$</p>
<p>Now, $$p - q + {q^2} = - 1 + \sqrt 3 + 3 = 2 + \sqrt 3 = \alpha $$</p>
<p>and $$p + q + {q^2} = - 1 - \sqrt 3 + 3 = 2 - \sqrt 3 = \beta $$</p>
<p>$$\therefore$$ $$\alpha + \beta = 4$$</p>
<p>$$\alpha \beta = \left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right) = 4 - 3 = 1$$</p>
<p>$$\therefore$$ Quadratic equation is</p>
<p>$${x^2} - (\alpha + \beta )x + \alpha \beta = 0$$</p>
<p>$$ \Rightarrow {x^2} - 4x + 1 = 0$$</p>
|
mcq
|
jee-main-2023-online-24th-january-morning-shift
|
UC6Gh9o9Zew8zAns
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
If $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$, then z lies on :
|
[{"identifier": "A", "content": "an ellipse"}, {"identifier": "B", "content": "the imaginary axis"}, {"identifier": "C", "content": "a circle"}, {"identifier": "D", "content": "the real axis"}]
|
["B"]
| null |
Given $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$,
<br><br>By squaring both sides we get,
<br><br>$${\left| {{z^2} - 1} \right|^2}$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$
<br><br>$$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\overline {\left( {{z^2} - 1} \right)} $$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$ [ as $${{{\left| z \right|}^2}}$$ = $$z\overline z $$ ]
<br><br>$$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\left( {{{\left( {\overline z } \right)}^2} - 1} \right)$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$
<br><br>$$ \Rightarrow $$ $${\left( {z\overline z } \right)^2}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1
<br><br>$$ \Rightarrow $$ $${\left| z \right|^4}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1
<br><br>$$ \Rightarrow $$ $${{z^2}}$$ $$+$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 2$${z\overline z }$$ = 0
<br><br>$$ \Rightarrow $$ $${\left( {z + \overline z } \right)^2}$$ = 0
<br><br>$$ \Rightarrow $$ $${z + \overline z }$$ = 0
<br><br>$$ \Rightarrow $$ $$z$$ = $$-$$ $${\overline z }$$
<br><br>If $$z$$ = x + iy
<br><br>then $${\overline z }$$ = x - iy
<br><br>$$\therefore$$ x + iy = - (x - iy)
<br><br>$$ \Rightarrow $$ x + iy = - x + iy
<br><br>$$ \Rightarrow $$ x = 0
<br><br>$$\therefore$$ z is purely imaginary.
<br><br>So, it is lie on the imaginary axis.
|
mcq
|
aieee-2004
|
de5FYCyGl5T5tbGQ
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
If $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$, then $$z$$ lies on :
|
[{"identifier": "A", "content": "an ellipse"}, {"identifier": "B", "content": "a circle"}, {"identifier": "C", "content": "a straight line"}, {"identifier": "D", "content": "a parabola"}]
|
["C"]
| null |
Given $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$
<br><br>$$\therefore$$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = \left| \omega \right|$$
<br><br>$$ \Rightarrow $$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = 1$$
<br><br>$$ \Rightarrow $$ $$\left| z \right| = \left| {z - {1 \over {\sqrt 3 }}i} \right|$$ ..........equation (1)
<br><br>$$\left| z \right|$$ represent distance of $$z$$ from point (0, 0) and
<br><br>$$\left| {z - {1 \over {\sqrt 3 }}i} \right|$$ represent distance of $$z$$ from point $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$.
<br><br>According to the equation (1) the distance of $$z$$ from point (0, 0) and $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$ is equal. Only if z is on a straight line then it will be equal distance from the both the points.
<img class="question-image" src="https://imagex.cdn.examgoal.net/9U9Fh0URP4SDcvKcK/MWBLvve80xYO19IP7fKY2TcwiUxoW/A4s0qXhuHL5WzzyZNFbBPy/image.svg" loading="lazy" alt="AIEEE 2005 Mathematics - Complex Numbers Question 157 English Explanation">
|
mcq
|
aieee-2005
|
HRDMy7FVhj3Qbxer
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
The number of complex numbers z such that $$\left| {z - 1} \right| = \left| {z + 1} \right| = \left| {z - i} \right|$$ equals :
|
[{"identifier": "A", "content": "1 "}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "$$\\infty $$ "}, {"identifier": "D", "content": "0 "}]
|
["A"]
| null |
Let $$z=x+iy$$
<br><br>$$\left| {z - 1} \right| = \left| {z + 1} \right|{\left( {x - 1} \right)^2} + {y^2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {x + 1} \right)^2} + {y^2}$$
<br><br>$$ \Rightarrow {\mathop{\rm Re}\nolimits} \,z = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x = 0$$
<br><br>$$\left| {z - 1} \right| = \left| {z - i} \right|{\left( {x - 1} \right)^2} + {y^2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}$$
<br><br>$$ \Rightarrow x = y$$
<br><br>$$\left| {z + 1} \right| = \left| {z - i} \right|{\left( {x + 1} \right)^2} + {y^2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}$$
<br><br>Only $$(0,0)$$ will satisfy all conditions.
<br><br>$$ \Rightarrow $$ Number of complex number $$z=1$$
|
mcq
|
aieee-2010
|
4bdMzc13up82nWp5
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
If z is a complex number such that $$\,\left| z \right| \ge 2\,$$, then the minimum value of $$\,\,\left| {z + {1 \over 2}} \right|$$ :
|
[{"identifier": "A", "content": "is strictly greater that $${{5 \\over 2}}$$"}, {"identifier": "B", "content": "is strictly greater that $${{3 \\over 2}}$$ but less than $${{5 \\over 2}}$$"}, {"identifier": "C", "content": "is equal to $${{5 \\over 2}}$$"}, {"identifier": "D", "content": "lie in the interval (1, 2)"}]
|
["B"]
| null |
We know minimum value of
<br><br>$$\,\,\,\left| {{Z_1} + {Z_2}} \right|\,\,\,$$ is $$\,\,\,\left| {\left| {{Z_1}} \right| - \left| {{Z_2}} \right|} \right|$$
<br><br>Thus minimum value of
<br><br>$$\,\,\,\left| {Z + {1 \over 2}} \right|\,\,\,$$ is $$\,\,\,\left| {\left| Z \right| - {1 \over 2}} \right|$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \le \left| {Z + {1 \over 2}} \right| \le \left| Z \right| + {1 \over 2}$$
<br><br>Since, $$\,\,\,\left| Z \right| \ge 2$$
<br><br>$$\therefore$$ $$\,\,\,2 - {1 \over 2} < \left| {Z + {1 \over 2}} \right| < 2 + {1 \over 2}$$
<br><br>$$ \Rightarrow {3 \over 2} < \left| {Z + {1 \over 2}} \right| < {5 \over 2}$$
|
mcq
|
jee-main-2014-offline
|
iSzQWpXnIqltJoVX
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
A complex number z is said to be unimodular if $$\,\left| z \right| = 1$$. Suppose $${z_1}$$ and $${z_2}$$ are complex numbers such that $${{{z_1} - 2{z_2}} \over {2 - {z_1}\overline {{z_2}} }}$$ is unimodular and $${z_2}$$ is not unimodular. Then the point $${z_1}$$ lies on a :
|
[{"identifier": "A", "content": "circle of radius 2."}, {"identifier": "B", "content": "circle of radius $${\\sqrt 2 }$$."}, {"identifier": "C", "content": "straight line parallel to x-axis"}, {"identifier": "D", "content": "straight line parallel to y-axis."}]
|
["A"]
| null |
$$\left| {{{{z_1} - 2{z_2}} \over {2 - {z_1}{{\overline z }_2}}}} \right| = 1 \Rightarrow {\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}{{\overline z }_2}} \right|^2}$$
<br><br>$$ \Rightarrow \left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1} - 2{z_2}} } \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {\overline {2 - {z_1}{{\overline z }_2}} } \right)$$
<br><br>$$ \Rightarrow \left( {{z_1} - 2{z_1}} \right)\left( {{{\overline z }_1} - 2{{\overline z }_2}} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left( {2 - {z_1}\overline {{z_2}} } \right)\left( {2 - {{\overline z }_1}{z_2}} \right)$$
<br><br>$$ \Rightarrow \left( {{z_1}{{\overline z }_1}} \right) - 2{z_1}{\overline z _2} - 2{\overline z _1}{z_2} + 4{z_2}{\overline z _2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 4 - 2{\overline z _1}{z_2} - 2{z_1}{\overline z _2} + {z_1}{\overline z _1}{z_2}{\overline z _2}$$
<br><br>$$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 4 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2}$$
<br><br>$$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} - 4 - {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} = 0$$
<br><br>$$\left( {{{\left| {{z_1}} \right|}^2} - 4} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0$$
<br><br>As $$\,\,\,\left| {{z_2}} \right| \ne 1\,\,\,$$ $$\therefore$$ $$\,\,\,{\left| {{z_1}} \right|^2} = 4 \Rightarrow \left| {{z_1}} \right| = 2$$
<br><br>$$ \Rightarrow \,$$ Point $$\,{z_1}\,$$ lies on circle of radius $$2.$$
|
mcq
|
jee-main-2015-offline
|
kVD6ruGEq9JlGdhtNQrqq
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
If |z $$-$$ 3 + 2i| $$ \le $$ 4 then the difference between the greatest value and the least value of |z| is :
|
[{"identifier": "A", "content": "$$2\\sqrt {13} $$"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "4 + $$\\sqrt {13} $$"}, {"identifier": "D", "content": "$$\\sqrt {13} $$"}]
|
["A"]
| null |
$$\left| {z - \left( {3 - 2i} \right)} \right| \le 4$$ represents a circle whose center is (3, $$-$$2) and radius = 4.
<br><br>$$\left| z \right|$$ = $$\left| z -0\right|$$ represents the distance of point 'z' from origin (0, 0)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266491/exam_images/mjmley3qs43gpsjau1nk.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 15th April Evening Slot Mathematics - Complex Numbers Question 140 English Explanation">
<br><br>Suppose RS is the normal of the circle passing through origin 'O' and G is its center (3, $$-$$2).
<br><br>Here, OR is the least distance
<br><br>and OS is in the greatest distance
<br><br>OR = RG $$-$$ OG and OS = OG + GS . . . . .(1)
<br><br>As, RG = GS = 4
<br><br>OG = $$\sqrt {{3^2} + \left( { - 2{)^2}} \right)} $$ = $$\sqrt {9 + 4} $$ = $$\sqrt {13} $$
<br><br>From (1), OR = 4 $$-$$ $$\sqrt {13} $$ and OS = 4 + $$\sqrt {13} $$
<br><br>So, required difference = $$\left( {4 + \sqrt {13} } \right)$$ $$-$$ $$\left( {4 - \sqrt {13} } \right)$$
<br><br>= $$\sqrt {13} + \sqrt {13} $$ = $$2\sqrt {13} $$
|
mcq
|
jee-main-2018-online-15th-april-evening-slot
|
l79mOqWrm4oKN8xh1VHfS
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
Let z<sub>1</sub> and z<sub>2</sub> be any two non-zero complex numbers such that $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|.$$ If $$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$ then :
|
[{"identifier": "A", "content": "$${\\rm I}m\\left( z \\right) = 0$$"}, {"identifier": "B", "content": "$$\\left| z \\right| = \\sqrt {{17 \\over 2}} $$"}, {"identifier": "C", "content": "$$\\left| z \\right| =$$ $${1 \\over 2}\\sqrt {9 + 16{{\\cos }^2}\\theta } $$"}, {"identifier": "D", "content": "Re(z) $$=$$ 0"}]
|
["C"]
| null |
Given, $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$$
<br><br>$$ \Rightarrow $$ $${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$$
<br><br>$$ \Rightarrow $$ $${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$$
<br><br>As we know, for any compled number
<br><br>$${{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}$$(cos$$\theta $$ + i sin$$\theta $$)
<br><br>= 2(cos$$\theta $$ + i sin$$\theta $$)
<br><br>$$ \therefore $$ $${{2{z_2}} \over {3{z_1}}}$$ = $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}$$
<br><br>= $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}$$
<br><br>= $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$
<br>Now, given
<br><br>$$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$
<br><br>= 2(cos$$\theta $$ + i sin$$\theta $$) + $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$
<br><br>= $${{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }$$
<br><br>So, |z| = $$\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta } $$
<br><br>= $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $$
<br><br>z is neither purely real nor purely imaginary and |z| depends on $$\theta $$.
|
mcq
|
jee-main-2019-online-10th-january-morning-slot
|
ogP6rmBUumLbGPZQG5z8i
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
Let z be a complex number such that |z| + z = 3 + i (where i = $$\sqrt { - 1} $$). Then |z| is equal to :
|
[{"identifier": "A", "content": "$${{\\sqrt {34} } \\over 3}$$"}, {"identifier": "B", "content": "$${5 \\over 3}$$"}, {"identifier": "C", "content": "$${5 \\over 4}$$"}, {"identifier": "D", "content": "$${{\\sqrt {41} } \\over 4}$$"}]
|
["B"]
| null |
$$\left| z \right| + z = 3 + i$$
<br><br>$$z = 3 - \left| z \right| + i$$
<br><br>Let $$3 - \left| z \right| = a \Rightarrow \left| z \right| = \left( {3 - a} \right)$$
<br><br>$$ \Rightarrow z = a + i \Rightarrow \left| z \right| = \sqrt {{a^2} + 1} $$
<br><br>$$ \Rightarrow 9 + {a^2} - 6a = {a^2} + 1 \Rightarrow a = {8 \over 6} = {4 \over 3}$$
<br><br>$$ \Rightarrow \left| z \right| = 3 - {4 \over 3} = {5 \over 3}$$
|
mcq
|
jee-main-2019-online-11th-january-evening-slot
|
LrLuK5pVFXqE5Ymzxn9Vn
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
Let z<sub>1</sub> and z<sub>2</sub> be two complex numbers satisfying | z<sub>1</sub> | = 9 and | z<sub>2</sub> – 3 – 4i | = 4. Then the minimum value of
| z<sub>1</sub> – z<sub>2</sub> | is :
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$\\sqrt 2 $$"}]
|
["A"]
| null |
$$\left| {{z_1}} \right| = 9,\,\,\left| {{z_2} - \left( {3 + 4i} \right)} \right| = 4$$
<br><br>$${C_1},(0,0)$$ radius r<sub>1</sub> = 9
<br><br>C<sub>2</sub> (3, 4), radius r<sub>2</sub> = 4
<br><br>C<sub>1</sub>C<sub>2</sub> = $$\left| {{r_1} - {r_2}} \right| = 5$$
<br><br>$$ \therefore $$ Circle touches internally
<br><br>$$ \therefore $$ $${\left| {{z_1} - {z_2}} \right|_{\min }} = 0$$
|
mcq
|
jee-main-2019-online-12th-january-evening-slot
|
5XxcQkY9Tu2ZkHWYwQ7k9k2k5ip2n7v
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
Let z be complex number such that
<br/> $$\left| {{{z - i} \over {z + 2i}}} \right| = 1$$ and |z| = $${5 \over 2}$$.<br/> Then the value of |z + 3i| is :
|
[{"identifier": "A", "content": "$$2\\sqrt 3 $$"}, {"identifier": "B", "content": "$$\\sqrt {10} $$"}, {"identifier": "C", "content": "$${{15} \\over 4}$$"}, {"identifier": "D", "content": "$${7 \\over 2}$$"}]
|
["D"]
| null |
Given $$\left| {{{z - i} \over {z + 2i}}} \right| = 1$$
<br><br>|z – i| = |z + 2i|
<br><br>(let z = x + iy)
<br><br>$$ \Rightarrow $$ x<sup>2</sup>
+ (y – 1)<sup>2</sup>
= x<sup>2</sup>
+ (y + 2)<sup>2</sup>
<br><br>$$ \Rightarrow $$ y = $$ - {1 \over 2}$$
<br><br>Also given |z| = $${5 \over 2}$$
<br><br>$$ \Rightarrow $$ x<sup>2</sup>
+ y<sup>2</sup> = $${{25} \over 4}$$
<br><br>$$ \Rightarrow $$ x<sup>2</sup> = 6
<br><br>$$ \therefore $$ z = $$ \pm \sqrt 6 $$ - $$ - {1 \over 2}i$$
<br><br>|z + 3i| = $$\sqrt {6 + {{25} \over 4}} $$ = $${7 \over 2}$$
|
mcq
|
jee-main-2020-online-9th-january-morning-slot
|
lOLmjWCdgMiaiJ2BWW7k9k2k5khn9k5
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
If z be a complex number satisfying
|Re(z)| + |Im(z)| = 4, then |z| cannot be :
|
[{"identifier": "A", "content": "$$\\sqrt {10} $$"}, {"identifier": "B", "content": "$$\\sqrt {7} $$"}, {"identifier": "C", "content": "$$\\sqrt {{{17} \\over 2}} $$"}, {"identifier": "D", "content": "$$\\sqrt {8} $$"}]
|
["B"]
| null |
Let z = x + iy
<br><br>given that |Re(z)| + |Im(z)| = 4
<br><br>$$ \therefore $$ |x| + |y| = 4
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264539/exam_images/tkloovrsmfpi6fjklioa.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 9th January Evening Slot Mathematics - Complex Numbers Question 114 English Explanation">
<br>Maximum value of |z| = 4
<br><br>Minimum value of |z| = perpendicular distance of line AB from (0, 0) = $$2\sqrt 2 $$
<br><br>$$ \therefore $$ |z| $$ \in $$ $$\left[ {2\sqrt 2 ,4} \right]$$
<br><br>$$ \therefore $$ |z| cannot be $$\sqrt {7} $$.
|
mcq
|
jee-main-2020-online-9th-january-evening-slot
|
peRrOzS3nUH7gT3DYS1klri4bcj
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
If the least and the largest real values of a, for which the <br/>equation z + $$\alpha $$|z – 1| + 2i = 0
(z $$ \in $$ C and i = $$\sqrt { - 1} $$) has a solution, are p and q respectively; then 4(p<sup>2</sup> + q<sup>2</sup>) is equal to __________.
|
[]
| null |
10
|
$$x + iy + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} + 2i = 0$$<br><br>$$ \therefore $$ y + 2 = 0 and $$x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0$$<br><br>y = $$-$$2 & $${x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4)$$<br><br>$${\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^2} + 5{\alpha ^2} = 0$$<br><br>$$x \in R \Rightarrow D \ge 0$$<br><br>$$4{\alpha ^4} - 4({\alpha ^2} - 1)5{\alpha ^2} \ge 0$$<br><br>$${\alpha ^2}[4{\alpha ^2} - 2{\alpha ^2} + 20] \ge 0$$<br><br>$${\alpha ^2}[ - 16{\alpha ^2} + 20] \ge 0$$<br><br>$${\alpha ^2}\left[ {{\alpha ^2} - {5 \over 4}} \right] \le 0$$<br><br>$$0 \le {\alpha ^2} \le {5 \over 4}$$<br><br>$$ \therefore $$ $${\alpha ^2} \in \left[ {0,{5 \over 4}} \right]$$<br><br>$$ \therefore $$ $$\alpha \in \left[ { - {{\sqrt 5 } \over 2},{{\sqrt 5 } \over 2}} \right]$$<br><br>then $$4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10$$
|
integer
|
jee-main-2021-online-24th-february-morning-slot
|
K1j14MVoLV2InNAjlE1kmhwyz0n
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
Let a complex number z, |z| $$\ne$$ 1, <br/><br/>satisfy $${\log _{{1 \over {\sqrt 2 }}}}\left( {{{|z| + 11} \over {{{(|z| - 1)}^2}}}} \right) \le 2$$. Then, the largest value of |z| is equal to ____________.
|
[{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "7"}]
|
["D"]
| null |
$${{|z| + 11} \over {{{(|z| - 1)}^2}}} \ge {1 \over 2}$$<br><br>$$2|z| + 22 \ge {(|z| - 1)^2}$$<br><br>$$2|z| + 22 \ge \,|z{|^2} - 2|z| + 1$$<br><br>$$|z{|^2} - 4|z| - 21 \le 0$$<br><br>$$(|z| - 7)(|z| + 3) \le 0$$<br><br>$$ \Rightarrow \,|z| \le 7$$<br><br>$$ \therefore $$ $$|z{|_{\max }} = 7$$
|
mcq
|
jee-main-2021-online-16th-march-morning-shift
|
ye1emd881Y8Z3i9nuW1kmiwzsy1
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
The least value of |z| where z is complex number which satisfies the inequality $$\exp \left( {{{(|z| + 3)(|z| - 1)} \over {||z| + 1|}}{{\log }_e}2} \right) \ge {\log _{\sqrt 2 }}|5\sqrt 7 + 9i|,i = \sqrt { - 1} $$, is equal to :
|
[{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$\\sqrt 5 $$"}]
|
["B"]
| null |
Let | z | = t, t $$ \ge $$ 0<br><br>$${e^{{{(t + 3)(t - 1)} \over {t + 1}}{{\log }_e}2}} \ge {\log _{\sqrt 2 }}16 = 8$$ ($$ \because $$ t + 1 > 0)<br><br>$${2^{{{(t + 3)(t - 1)} \over {t + 1}}}} \ge {2^3}$$<br><br>$${{(t + 3)(t - 1)} \over {t + 1}} \ge 3$$<br><br>$${t^2} + 2t - 3 \ge 3t + 3$$<br><br>$${t^2} - t - 6 \ge 0$$<br><br>$$t \in ( - \infty , - 2) \cup [3,\infty )$$ But t $$ \ge $$ 0<br><br>$$ \therefore $$ $$t \in [3,\infty )$$
|
mcq
|
jee-main-2021-online-16th-march-evening-shift
|
1ktk9aapu
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
If z is a complex number such that $${{z - i} \over {z - 1}}$$ is purely imaginary, then the minimum value of | z $$-$$ (3 + 3i) | is :
|
[{"identifier": "A", "content": "$$2\\sqrt 2 - 1$$"}, {"identifier": "B", "content": "$$3\\sqrt 2 $$"}, {"identifier": "C", "content": "$$6\\sqrt 2 $$"}, {"identifier": "D", "content": "$$2\\sqrt 2 $$"}]
|
["D"]
| null |
$${{z - i} \over {z - 1}}$$ is purely imaginary number<br><br>Let $$z = x + iy$$<br><br>$$\therefore$$ $${{x + i(y - 1)} \over {(x - 1) + i(y)}} \times {{(x - 1) - iy} \over {(x - 1) - iy}}$$<br><br>$$ \Rightarrow {{x(x - 1) + y(y - 1) + i( - y - x + 1)} \over {{{(x - 1)}^2} + {y^2}}}$$ is purely imaginary number<br><br>$$ \Rightarrow x(x - 1) + y(y - 1) = 0$$<br><br>$$ \Rightarrow {\left( {x - {1 \over 2}} \right)^2} + {\left( {y - {1 \over 2}} \right)^2} = {1 \over 2}$$<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263291/exam_images/u9prdssniktivaw2yk0n.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267396/exam_images/obzlgxtqwruolvxo2ezb.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263710/exam_images/bbjacsejtmgqpyhao2r7.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264898/exam_images/ry4oynpllxb7fvr8cjyv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Evening Shift Mathematics - Complex Numbers Question 75 English Explanation"></picture> <br><br>$$\therefore$$ $${\left| {z - (3 + 3i)} \right|_{\min }} = \left| {PC} \right| - {1 \over {\sqrt 2 }}$$<br><br>$$ = {5 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 2\sqrt 2 $$
|
mcq
|
jee-main-2021-online-31st-august-evening-shift
|
1ktoaywlm
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
If for the complex numbers z satisfying | z $$-$$ 2 $$-$$ 2i | $$\le$$ 1, the maximum value of | 3iz + 6 | is attained at a + ib, then a + b is equal to ______________.
|
[]
| null |
5
|
| z $$-$$ 2 $$-$$ 2i | $$\le$$ 1<br><br>| x + iy $$-$$ 2 $$-$$ 2i | $$\le$$ 1<br><br>|(x $$-$$ 2) + i(y $$-$$ 2)| $$\le$$ 1<br><br>(x $$-$$ 2)<sup>2</sup> + (y $$-$$ 2)<sup>2</sup> $$\le$$ 1<br><br>| 3iz + 6 |<sub>max</sub> at a + ib<br><br>$$\left| {3i} \right|\left| {z + {6 \over {3i}}} \right|$$<br><br>$$3{\left| {z - 2i} \right|_{\max }}$$<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwoq4x6f/6cf72af3-199a-4296-97de-3bb71dc1f158/c6409070-534d-11ec-9cbb-695a838b20fb/file-1kwoq4x6g.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kwoq4x6f/6cf72af3-199a-4296-97de-3bb71dc1f158/c6409070-534d-11ec-9cbb-695a838b20fb/file-1kwoq4x6g.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 1st September Evening Shift Mathematics - Complex Numbers Question 76 English Explanation"><br>From figure maximum distance at 3 + 2i<br><br>a + ib = 3 + 2i = a + b = 3 + 2 = 5 Ans.
|
integer
|
jee-main-2021-online-1st-september-evening-shift
|
1l5bay6xr
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>Let S = {z $$\in$$ C : |z $$-$$ 3| $$\le$$ 1 and z(4 + 3i) + $$\overline z $$(4 $$-$$ 3i) $$\le$$ 24}. If $$\alpha$$ + i$$\beta$$ is the point in S which is closest to 4i, then 25($$\alpha$$ + $$\beta$$) is equal to ___________.</p>
|
[]
| null |
80
|
<p>Here $$|z - 3| < 1$$</p>
<p>$$ \Rightarrow {(x - 3)^2} + {y^2} < 1$$</p>
<p>and $$z = (4 + 3i) + \overline z (4 - 3i) \le 24$$</p>
<p>$$ \Rightarrow 4x - 3y \le 12$$</p>
<p>$$\tan \theta = {4 \over 3}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5v605na/b9bf1f33-cb41-4f86-92e8-8094f98d7ae7/7a1f0360-0906-11ed-a790-b11fa70c8a36/file-1l5v605nb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5v605na/b9bf1f33-cb41-4f86-92e8-8094f98d7ae7/7a1f0360-0906-11ed-a790-b11fa70c8a36/file-1l5v605nb.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Evening Shift Mathematics - Complex Numbers Question 63 English Explanation"></p>
<p>$$\therefore$$ Coordinate of $$P = (3 - \cos \theta ,\sin \theta )$$</p>
<p>$$ = \left( {3 - {3 \over 5},{4 \over 5}} \right)$$</p>
<p>$$\therefore$$ $$\alpha + i\beta = {{12} \over 5} + {4 \over 5}i$$</p>
<p>$$\therefore$$ $$25(\alpha + \beta ) = 80$$</p>
|
integer
|
jee-main-2022-online-24th-june-evening-shift
|
1l5c0riod
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>Let $$A = \{ z \in C:1 \le |z - (1 + i)| \le 2\} $$</p>
<p>and $$B = \{ z \in A:|z - (1 - i)| = 1\} $$. Then, B :</p>
|
[{"identifier": "A", "content": "is an empty set"}, {"identifier": "B", "content": "contains exactly two elements"}, {"identifier": "C", "content": "contains exactly three elements"}, {"identifier": "D", "content": "is an infinite set"}]
|
["D"]
| null |
<p>Let, $$z = x + iy$$</p>
<p>Given, $$1 \le \left| {z - (1 + i)} \right| \le 2$$</p>
<p>$$ \Rightarrow 1 \le \left| {x + iy - 1 - i} \right| \le 2$$</p>
<p>$$ \Rightarrow 1 \le \left| {(x - 1) + i(y - 1)} \right| \le 2$$</p>
<p>$$ \Rightarrow 1 \le \sqrt {{{(x - 1)}^2} + {{(y - 1)}^2}} \le 2$$</p>
<p>It represent two concentric circle both have center at (1, 1) and radius 1 and 2.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5sd6b1c/1293a42c-142d-44c5-b0b7-b47b530256fa/2b18ad10-077c-11ed-94f6-83604aa63acb/file-1l5sd6b1d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5sd6b1c/1293a42c-142d-44c5-b0b7-b47b530256fa/2b18ad10-077c-11ed-94f6-83604aa63acb/file-1l5sd6b1d.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Morning Shift Mathematics - Complex Numbers Question 62 English Explanation 1"></p>
<p>Also given,</p>
<p>$$\left| {z - (1 - i)} \right| = 1$$</p>
<p>$$ \Rightarrow \left| {x + iy - 1 + i} \right| = 1$$</p>
<p>$$ \Rightarrow \left| {(x - 1) + i(y + 1)} \right| = 1$$</p>
<p>$$ \Rightarrow \sqrt {{{(x - 1)}^2} + {{(y + 1)}^2}} = 1$$</p>
<p>This represent a circle with center at (1, $$-$$1) and radius = 1.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5scxr6i/1a1381e8-31b3-4230-9271-a7ce070fdff2/3d4976f0-077b-11ed-94f6-83604aa63acb/file-1l5scxr6n.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5scxr6i/1a1381e8-31b3-4230-9271-a7ce070fdff2/3d4976f0-077b-11ed-94f6-83604aa63acb/file-1l5scxr6n.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Morning Shift Mathematics - Complex Numbers Question 62 English Explanation 2"></p>
<p>In the common region infinite values of B possible.</p>
|
mcq
|
jee-main-2022-online-24th-june-morning-shift
|
1l6duxx3p
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>For $$\mathrm{n} \in \mathbf{N}$$, let $$\mathrm{S}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-3+2 i|=\frac{\mathrm{n}}{4}\right\}$$ and $$\mathrm{T}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-2+3 i|=\frac{1}{\mathrm{n}}\right\}$$. Then the number of elements in the set $$\left\{n \in \mathbf{N}: S_{n} \cap T_{n}=\phi\right\}$$ is :
</p>
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}]
|
["D"]
| null |
$S_{n}=\left\{z \in C:|z-3+2 i|=\frac{n}{4}\right\}$ represents a circle with centre $C_{1}(3,-2)$ and radius $r_{1}=\frac{n}{4}$
<br/><br/>
Similarly $T_{n}$ represents circle with centre $C_{2}(2,-3)$ and radius $r_{2}=\frac{1}{n}$
<br/><br/>
$\text { As } S_{n} \cap T_{n}=\phi$<br/><br/>
$$
\begin{array}{llr}
C_{1} C_{2}>r_{1}+r_{2}& \text { OR } & C_{1} C_{2}<\left|r_{1}-r_{2}\right| \\\\
\sqrt{2}>\frac{n}{4}+\frac{1}{n} & \text { OR } & \sqrt{2}<\left|\frac{n}{4}-\frac{1}{n}\right| \\\\
n=1,2,3,4&& n \text { may take infinite values }
\end{array}
$$
|
mcq
|
jee-main-2022-online-25th-july-morning-shift
|
1l6f0l8mz
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>For $$z \in \mathbb{C}$$ if the minimum value of $$(|z-3 \sqrt{2}|+|z-p \sqrt{2} i|)$$ is $$5 \sqrt{2}$$, then a value Question: of $$p$$ is _____________.</p>
|
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$\\frac{7}{2}$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$\\frac{9}{2}$$"}]
|
["C"]
| null |
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7bug80c/85c5d49d-7e9c-43b0-b928-429394532f1d/d5c2fcc0-25fe-11ed-b97b-f3de74335fe6/file-1l7bug80d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7bug80c/85c5d49d-7e9c-43b0-b928-429394532f1d/d5c2fcc0-25fe-11ed-b97b-f3de74335fe6/file-1l7bug80d.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Evening Shift Mathematics - Complex Numbers Question 59 English Explanation"></p>
<p>It is sum of distance of z from $$\left( {3\sqrt 2 ,0} \right)$$ and $$\left( {0,p\sqrt 2 } \right)$$</p>
<p>For minimising, z should lie on AB and $$AB = 5\sqrt 2 $$</p>
<p>$${(AB)^2} = 18 + 2{p^2}$$</p>
<p>$$p = \, \pm \,4$$</p>
|
mcq
|
jee-main-2022-online-25th-july-evening-shift
|
1l6hxiqss
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>If $$z=x+i y$$ satisfies $$|z|-2=0$$ and $$|z-i|-|z+5 i|=0$$, then :</p>
|
[{"identifier": "A", "content": "$$x+2 y-4=0$$"}, {"identifier": "B", "content": "$$x^{2}+y-4=0$$"}, {"identifier": "C", "content": "$$x+2 y+4=0$$"}, {"identifier": "D", "content": "$$x^{2}-y+3=0$$"}]
|
["C"]
| null |
<p>$$|z - i| = |z + 5i|$$</p>
<p>So, $$\mathrm{z}$$ lies on $${ \bot ^r}$$ bisector of $$(0,1)$$ and $$(0, - 5)$$</p>
<p>i.e., line $$y = - 2$$</p>
<p>as $$|z| = 2$$</p>
<p>$$ \Rightarrow z = - 2i$$</p>
<p>$$x = 0$$ and $$y = - 2$$</p>
<p>so, $$x + 2y + 4 = 0$$</p>
|
mcq
|
jee-main-2022-online-26th-july-evening-shift
|
1l6jb3png
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>Let the minimum value $$v_{0}$$ of $$v=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \in \mathbb{C}$$ is attained at $${ }{z}=z_{0}$$. Then $$\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}$$ is equal to :</p>
|
[{"identifier": "A", "content": "1000"}, {"identifier": "B", "content": "1024"}, {"identifier": "C", "content": "1105"}, {"identifier": "D", "content": "1196"}]
|
["A"]
| null |
<p>Let $$z = x + iy$$</p>
<p>$$v = {x^2} + {y^2} + {(x - 3)^2} + {y^2} + {x^2} + {(y - 6)^2}$$</p>
<p>$$ = (3{x^2} - 6x + 9) + (3{y^2} - 12y + 36)$$</p>
<p>$$ = 3({x^2} + {y^2} - 2x - 4y + 15)$$</p>
<p>$$ = 3[{(x - 1)^2} + {(y - 2)^2} + 10]$$</p>
<p>$${v_{\min }}$$ at $$z = 1 + 2i = {z_0}$$ and $${v_0} = 30$$</p>
<p>so $$|2{(1 + 2i)^2} - {(1 - 2i)^3} + 3{|^2} + 900$$</p>
<p>$$ = |2( - 3 + 4i) - (1 - 8{i^3} - 6i(1 - 2i) + 3{|^2} + 900$$</p>
<p>$$ = | - 6 + 8i - (1 + 8i - 6i - 12) + 3{|^2} + 900$$</p>
<p>$$ = |8 + 6i{|^2} + 900$$</p>
<p>$$ = 1000$$</p>
|
mcq
|
jee-main-2022-online-27th-july-morning-shift
|
1l6m60fcu
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>Let $$S_{1}=\left\{z_{1} \in \mathbf{C}:\left|z_{1}-3\right|=\frac{1}{2}\right\}$$ and $$S_{2}=\left\{z_{2} \in \mathbf{C}:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\}$$. Then, for $$z_{1} \in S_{1}$$ and $$z_{2} \in S_{2}$$, the least value of $$\left|z_{2}-z_{1}\right|$$ is :</p>
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$\\frac{1}{2}$$"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$"}, {"identifier": "D", "content": "$$\\frac{5}{2}$$"}]
|
["C"]
| null |
<p>$$\because$$ $${\left| {{Z_2} + |{Z_2} - 1|} \right|^2} = {\left| {{Z_2} - |{Z_2} + 1|} \right|^2}$$</p>
<p>$$ \Rightarrow \left( {{Z_2} + |{Z_2} - 1|} \right)\left( {{{\overline Z }_2} + |{Z_2} - 1|} \right) = \left( {{Z_2} - |{Z_2} + 1|} \right)\left( {{{\overline Z }_2} - |{Z_2} + 1|} \right)$$</p>
<p>$$ \Rightarrow {Z_2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) + {\overline Z _2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) = |{Z_2} + 1{|^2} - |{Z_2} - 1{|^2}$$</p>
<p>$$ \Rightarrow \left( {{Z_2} + {{\overline Z }_2}} \right)\left( {|{Z_2} + 1| + |{Z_2} - 1|} \right) = 2\left( {{Z_2} + {{\overline Z }_2}} \right)$$</p>
<p>$$\Rightarrow$$ Either $${Z_2} + {\overline Z _2} = 0$$ or $$|{Z_2} + 1| + |{Z_2} - 1| = 2$$</p>
<p>So, Z<sub>2</sub> lies on imaginary axis or on real axis within $$[ - 1,1]$$</p>
<p>Also $$|{Z_1} - 3| = {1 \over 2} \Rightarrow {Z_1}$$ lies on the circle having center 3 and radius $${1 \over 2}$$.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb1odr/f03d7ef3-d4f9-4323-b266-e49dead29829/7e3573f0-2e7f-11ed-8702-156c00ced081/file-1l7rb1ods.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7rb1odr/f03d7ef3-d4f9-4323-b266-e49dead29829/7e3573f0-2e7f-11ed-8702-156c00ced081/file-1l7rb1ods.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th July Morning Shift Mathematics - Complex Numbers Question 52 English Explanation"></p>
<p>Clearly $$|{Z_1} - {Z_2}{|_{\min }} = {3 \over 2}$$</p>
|
mcq
|
jee-main-2022-online-28th-july-morning-shift
|
1l6rdnyvc
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>If $$z \neq 0$$ be a complex number such that $$\left|z-\frac{1}{z}\right|=2$$, then the maximum value of $$|z|$$ is :</p>
|
[{"identifier": "A", "content": "$$\\sqrt{2}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\sqrt{2}-1$$"}, {"identifier": "D", "content": "$$\\sqrt{2}+1$$"}]
|
["D"]
| null |
<p>We know,</p>
<p>$$\left| {|{z_1}| - |{z_2}|} \right| \le \left| {{z_1} + {z_2}} \right| \le |{z_1}| + |{z_2}|$$</p>
<p>$$\therefore$$ $$\left| {|z| - {1 \over {|z|}}} \right| \le \left| {z - {1 \over z}} \right|$$</p>
<p>$$ \Rightarrow \left| {|z| - {1 \over {|z|}}} \right| \le 2$$ [Given $$\left| {z - {1 \over z}} \right| = 2$$]</p>
<p>$$ \Rightarrow \left| {{{|z{|^2} - 1} \over {|z|}}} \right| \le 2$$</p>
<p>$$ \Rightarrow - 2 \le {{|z{|^2} - 1} \over {|z|}} \le 2$$</p>
<p>$$\therefore$$ $${{|z{|^2} - 1} \over {|z|}} \le 2$$</p>
<p>$$ \Rightarrow |z{|^2} - 1 \le 2|z|$$</p>
<p>$$ \Rightarrow |z{|^2} - 2|z| - 1 \le 0$$</p>
<p>$$ \Rightarrow |z{|^2} - 2|z| + 1 - 2 \le 0$$</p>
<p>$$ \Rightarrow {(|z| - 1)^2} - 2 \le 0$$</p>
<p>$$ \Rightarrow - \sqrt 2 \le |z| - 1 \le \sqrt 2 $$</p>
<p>$$ \Rightarrow 1 - \sqrt 2 \le |z| \le 1 + \sqrt 2 $$ ..... (1)</p>
<p>or</p>
<p>$$ - 2 \le {{|z{|^2} - 1} \over {|z|}}$$</p>
<p>$$ \Rightarrow |z{|^2} - 1 \le - 2|z|$$</p>
<p>$$ \Rightarrow |z{|^2} + 2|z| - 1 \le 0$$</p>
<p>$$ \Rightarrow |z{|^2} + 2|z| + 1 - 2 \le 0$$</p>
<p>$$ \Rightarrow {(|z| + 1)^2} - 2 \le 0$$</p>
<p>$$ \Rightarrow - \sqrt 2 \le |z| + 1 \le + \,\sqrt 2 $$</p>
<p>$$ \Rightarrow - \sqrt 2 - 1 \le |z| \le \sqrt 2 - 1$$ ...... (2)</p>
<p>From (1) and (2) we get,</p>
<p>Maximum value of $$|z| = \sqrt 2 + 1$$ and minimum value of $$|z| = - \sqrt 2 - 1$$</p>
|
mcq
|
jee-main-2022-online-29th-july-evening-shift
|
1ldu5cviu
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>Let $$z$$ be a complex number such that $$\left| {{{z - 2i} \over {z + i}}} \right| = 2,z \ne - i$$. Then $$z$$ lies on the circle of radius 2 and centre :</p>
|
[{"identifier": "A", "content": "(0, $$-$$2)"}, {"identifier": "B", "content": "(0, 0)"}, {"identifier": "C", "content": "(0, 2)"}, {"identifier": "D", "content": "(2, 0)"}]
|
["A"]
| null |
$\left|\frac{z-2 i}{z+i}\right|=2$
<br/><br/>
$\Rightarrow (z-2 i)(\bar{z}+2 i)=4(z+i)(\bar{z}-i)$
<br/><br/>
$\Rightarrow z \bar{z}+2 i z-2 i \bar{z}+4=4(z \bar{z}-z i+\overline{z i}+1)$
<br/><br/>
$\Rightarrow 3 z \bar{z}-6 i z+6 i \bar{z}=0$
<br/><br/>
$\Rightarrow z \bar{z}-2 i z+2 i \bar{z}=0$
<br/><br/>
$\therefore$ Centre $(-2 i)$ or $(0,-2)$
<br/><br/><b>Other Method :</b>
<br/><br/>$\left|\frac{z-2 i}{z+i}\right|=2, z \neq-i$
<br/><br/>Put $z=x+i y$
<br/><br/>$$
\begin{aligned}
& \left|\frac{x+i y-2 i}{x+i y+i}\right|=2 \\\\
& \Rightarrow \left|\frac{x+i(y-2)}{x+i(y+1)}\right|^2=4 \\\\
& \Rightarrow x^2+(y-2)^2=4\left[x^2+(y+1)^2\right] \\\\
& \Rightarrow x^2+y^2+4-4 y=4\left[x^2+y^2+1+2 y\right]
\end{aligned}
$$
<br/><br/>$\Rightarrow$ $x^2+y^2+4 y=0$
<br/><br/>or $x^2+(y+2)^2=2^2$
<br/><br/>$$ \therefore $$ Centre is $(0,-2)$.
|
mcq
|
jee-main-2023-online-25th-january-evening-shift
|
1lgzzx8yo
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>If for $$z=\alpha+i \beta,|z+2|=z+4(1+i)$$, then $$\alpha+\beta$$ and $$\alpha \beta$$ are the roots of the equation :</p>
|
[{"identifier": "A", "content": "$$x^{2}+2 x-3=0$$"}, {"identifier": "B", "content": "$$x^{2}+3 x-4=0$$"}, {"identifier": "C", "content": "$$x^{2}+x-12=0$$"}, {"identifier": "D", "content": "$$x^{2}+7 x+12=0$$"}]
|
["D"]
| null |
Given : $|z+2|=z+4(1+i)$
<br/><br/>Also, $z=\alpha+i \beta$
<br/><br/>$$
\begin{aligned}
& \therefore|z+2|=|\alpha+i \beta+2|=(\alpha+i \beta)+4+4 i \\\\
& \Rightarrow|(\alpha+2)+i \beta|=(\alpha+4)+i(\beta+4) \\\\
& \Rightarrow \sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+i(\beta+4) \\\\
& \Rightarrow \beta+4=0 \Rightarrow \beta=-4
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Now, }(\alpha+2)^2+\beta^2=(\alpha+4)^2 \\\\
& \Rightarrow \alpha^2+4+4 \alpha+\beta^2=\alpha^2+16+8 \alpha \\\\
& \Rightarrow 4+4 \alpha+16=16+8 \alpha \\\\
& \Rightarrow 4 \alpha=4 \Rightarrow \alpha=1 \\\\
& \text { So, } \alpha+\beta=-3 \text { and } \alpha \beta=-4 \\\\
& \therefore \text { Required equation is } \\\\
& x^2-(-3-4) x+(-3)(-4)=0 \\\\
& \Rightarrow x^2+7 x+12=0
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-8th-april-morning-shift
|
lsam3k7w
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
If $z$ is a complex number such that $|z| \leqslant 1$, then the minimum value of $\left|z+\frac{1}{2}(3+4 i)\right|$ is :
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$\\frac{5}{2}$"}, {"identifier": "C", "content": "$\\frac{3}{2}$"}, {"identifier": "D", "content": "3"}]
|
["C"]
| null |
<p>To find the minimum value of $\left|z+\frac{1}{2}(3+4i)\right|$, where $z$ is a complex number with $|z| \leqslant 1$, we can think of this geometrically as the distance from any point inside or on the boundary of the unit circle in the complex plane to the fixed point $\frac{1}{2}(3+4i)$.</p>
<p>To make this more clear, first write the expression as follows:
$$\left|z+\frac{1}{2}(3+4i)\right| = \left|z+\frac{3}{2}+\frac{4}{2}i\right| = \left|z+1.5+2i\right|$$</p>
<p>This means we are looking for the distance between any point on the complex plane represented by $z$ (where $z$ has a magnitude of up to 1) and the point $1.5+2i$.</p>
<p>The triangle inequality gives us the following relationship for any two complex numbers $z_1$ and $z_2$:
$$\left|z_1 + z_2\right| \geqslant \left| \left|z_1\right| - \left|z_2\right| \right|$$</p>
<p>Applying this to our specific case ($z_1 = z$ and $z_2 = -1.5 - 2i$):
$$\left|z + 1.5 + 2i\right| \geqslant \left| \left|z\right| - \left|-1.5 - 2i\right| \right|$$</p>
<p>Since $z$ is within the unit circle, $|z| \leqslant 1$ and $|-1.5 - 2i| = \sqrt{1.5^2 + 2^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5$.</p>
<p>So now we have:
$$\left|z + 1.5 + 2i\right| \geqslant \left| 1 - 2.5 \right|$$</p>
<p>Which simplifies to:
$$\left|z + 1.5 + 2i\right| \geqslant 1.5$$</p>
<p>Therefore, the minimum value of $\left|z+\frac{1}{2}(3+4i)\right|$ given that $|z| \leqslant 1$ is $\frac{3}{2}$.</p>
<p>So the correct answer is:</p>
<p>Option C: $\frac{3}{2}$</p>
|
mcq
|
jee-main-2024-online-1st-february-evening-shift
|
luy9clkm
|
maths
|
complex-numbers
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modulus-of-complex-numbers
|
<p>The sum of the square of the modulus of the elements in the set $$\{z=\mathrm{a}+\mathrm{ib}: \mathrm{a}, \mathrm{b} \in \mathbf{Z}, z \in \mathbf{C},|z-1| \leq 1,|z-5| \leq|z-5 \mathrm{i}|\}$$ is __________.</p>
|
[]
| null |
9
|
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw3in2th/8eab885e-7362-426e-8f81-a5a9157aded7/ca24c350-105b-11ef-9330-f7b57aaea8e9/file-1lw3in2ti.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw3in2th/8eab885e-7362-426e-8f81-a5a9157aded7/ca24c350-105b-11ef-9330-f7b57aaea8e9/file-1lw3in2ti.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Complex Numbers Question 9 English Explanation"></p>
<p>$$z$$ should be lying in the shaded region shown in adjacent figure.</p>
<p>Possible $$z$$ are</p>
<p>$$z=0+0 i,(1+0 i),(2+0 i),(1-i),(1+i)$$</p>
<p>Sum of squares of modulus</p>
<p>$$\begin{aligned}
& =0+1+4+2+2 \\
& =9
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-9th-april-morning-shift
|
lv7v3k13
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>Consider the following two statements :</p>
<p>Statement I: For any two non-zero complex numbers $$z_1, z_2,(|z_1|+|z_2|)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \text {, and }$$</p>
<p>Statement II : If $$x, y, z$$ are three distinct complex numbers and $$\mathrm{a}, \mathrm{b}, \mathrm{c}$$ are three positive real numbers such that $$\frac{\mathrm{a}}{|y-z|}=\frac{\mathrm{b}}{|z-x|}=\frac{\mathrm{c}}{|x-y|}$$, then $$\frac{\mathrm{a}^2}{y-z}+\frac{\mathrm{b}^2}{z-x}+\frac{\mathrm{c}^2}{x-y}=1$$.</p>
<p>Between the above two statements,</p>
|
[{"identifier": "A", "content": "both Statement I and Statement II are incorrect.\n"}, {"identifier": "B", "content": "Statement I is correct but Statement II is incorrect.\n"}, {"identifier": "C", "content": "Statement I is incorrect but Statement II is correct.\n"}, {"identifier": "D", "content": "both Statement I and Statement II are correct."}]
|
["B"]
| null |
<p>$$\begin{aligned}
& \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}=\lambda \\
& \Rightarrow a^2=\lambda^2|(y-z)|^2 \\
& b^2=\lambda^2|(z-x)|^2 \\
& c^2=\lambda^2|(x-y)|^2 \\
& \frac{a^2(\overline{y-z})}{(y-z)(y-z)}=\frac{a^2(\bar{y}-\bar{z})}{|y-z|^2}=\frac{a^2(\bar{y}-\bar{z})}{\frac{a^2}{\lambda^2}}=\lambda^2(\bar{y}-\bar{z}) \\
& \Rightarrow \sum\left(\frac{a^2}{y-z}\right)=\lambda^2(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0 \neq 1 \\
\end{aligned}$$</p>
<p>Statement I</p>
<p>$$\begin{aligned}
& \left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \\
& \Rightarrow \quad z_1=\left|z_1\right| e^{i \theta_1} \\
& \quad z_2=\left|z_2\right| e^{i \theta_2} \\
& \Rightarrow \frac{z_1}{\left|z_1\right|}=e^{i \theta_1} \\
& \Rightarrow \frac{z_2}{\left|z_2\right|}=e^{i \theta_2} \\
& \Rightarrow\left|e^{i \theta_1}+e^{i \theta_2}\right| \\
& \quad=\left|\sqrt{2+2 \cos \left(\theta_1-\theta_2\right)}\right| \\
& \quad=\left|2 \cos \left(\frac{\theta_1-\theta_2}{2}\right)\right| \leq 2
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-5th-april-morning-shift
|
lvb2948r
|
maths
|
complex-numbers
|
modulus-of-complex-numbers
|
<p>If $$z_1, z_2$$ are two distinct complex number such that $$\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2$$, then</p>
|
[{"identifier": "A", "content": "either $$z_1$$ lies on a circle of radius $$\\frac{1}{2}$$ or $$z_2$$ lies on a circle of radius 1.\n"}, {"identifier": "B", "content": "$$z_1$$ lies on a circle of radius $$\\frac{1}{2}$$ and $$z_2$$ lies on a circle of radius 1.\n"}, {"identifier": "C", "content": "either $$z_1$$ lies on a circle of radius 1 or $$z_2$$ lies on a circle of radius $$\\frac{1}{2}$$.\n"}, {"identifier": "D", "content": "both $$z_1$$ and $$z_2$$ lie on the same circle."}]
|
["C"]
| null |
<p>$$\begin{aligned}
& \left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2 \\
& \left|z_1-2 z_2\right|=\left|1-2 z_1 \bar{z}_2\right| \\
& \Rightarrow\left(z_1-2 z_2\right)\left(\bar{z}_1-2 \bar{z}_2\right)=\left(1-2 z_1 \bar{z}_2\right)\left(1-2 \bar{z}_1 z_2\right) \\
& \Rightarrow\left|z_1\right|^2+4\left|z_2\right|^2-2 \bar{z}_1 z_2-2 \bar{z}_2 z_1 \\
& \quad=1+4\left|z_1\right|^2\left|z_2\right|^2-2 z_1 \bar{z}_2-2 \bar{z}_1 z_2 \\
& \Rightarrow\left|z_1\right|^2+4\left|z_2\right|^2-4\left|z_1\right|^2\left|z_2\right|^2-1=0 \\
& \Rightarrow\left(\left|z_1\right|^2-1\right)\left(1-4\left|z_2\right|^2\right)=0 \\
& \Rightarrow\left|z_1\right|=1 \text { and }\left|z_2\right|=\frac{1}{2}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-6th-april-evening-shift
|
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