Complex Numbers
206 Questions
Start JEE Mains Test
2022
Q101
JEE Mains
MCQ
14 Mar 2026
The area of the polygon, whose vertices are the non-real roots of the equation $\overline z = i{z^2}$ is :
A.
${{3\sqrt 3 } \over 4}$
B.
${{3\sqrt 3 } \over 2}$
C.
${3 \over 2}$
D.
${3 \over 4}$
2022
Q102
JEE Mains
MCQ
14 Mar 2026
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 :
A.
a portion of a circle centred at $\left( {0, - {1 \over {\sqrt 3 }}} \right)$ that lies in the second and third quadrants only
B.
a portion of a circle centred at $\left( {0, - {1 \over {\sqrt 3 }}} \right)$ that lies in the second quadrant only
C.
an empty
D.
a portion of a circle of radius ${2 \over {\sqrt 3 }}$ that lies in the third quadrant only
2022
Q103
JEE Mains
MCQ
14 Mar 2026
Let z1 and z2 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 :
A.
$\arg {z_2} = {\pi \over 4}$
B.
$\arg {z_2} = - {{3\pi } \over 4}$
C.
$\arg {z_1} = {\pi \over 4}$
D.
$\arg {z_1} = - {{3\pi } \over 4}$
2022
Q104
JEE Mains
MCQ
14 Mar 2026
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 z1 is perpendicular to the line through z2 and z3, then $arg(z)$ is equal to :
A.
${\tan ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right) - \pi $
B.
${\tan ^{ - 1}}\left( {{{24} \over 7}} \right) - \pi $
C.
${\tan ^{ - 1}}\left( 3 \right) - \pi $
D.
${\tan ^{ - 1}}\left( {{3 \over 4}} \right) - \pi $
2022
Q105
JEE Mains
MCQ
14 Mar 2026
Let $A = \{ z \in C:1 \le |z - (1 + i)| \le 2\} $
and $B = \{ z \in A:|z - (1 - i)| = 1\} $. Then, B :
A.
is an empty set
B.
contains exactly two elements
C.
contains exactly three elements
D.
is an infinite set
2022
Q106
JEE Mains
Numerical
14 Mar 2026
Let $\mathrm{z}=a+i b, b \neq 0$ be complex numbers satisfying $z^{2}=\bar{z} \cdot 2^{1-z}$. Then the least value of $n \in N$, such that $z^{n}=(z+1)^{n}$, is equal to __________.
Correct Answer: 6
Explanation:
$\because$ ${z^2} = \overline z \,.\,{2^{1 - |z|}}$ ...... (1)
$ \Rightarrow |z{|^2} = |\overline z |\,.\,{2^{1 - |z|}}$
$ \Rightarrow |z| = {2^{1 - |z|}}$,
$\because$ $b \ne 0 \Rightarrow |z| \ne 0$
$\therefore$ $|z| = 1$ ...... (2)
$\because$ $z = a + ib$ then $\sqrt {{a^2} + {b^2}} = 1$ ...... (3)
Now again from equation (1), equation (2), equation (3) we get :
${a^2} - {b^2} + i2ab = (a - ib){2^0}$
$\therefore$ ${a^2} - {b^2} = a$ and $2ab = - b$
$\therefore$ $a = - {1 \over 2}$ and $b = \, \pm \,{{\sqrt 3 } \over 2}$
$z = - {1 \over 2} + {{\sqrt 3 } \over 2}i$ or $z = - {1 \over 2} - {{\sqrt 3 } \over 2}i$
${z^n} = {(z + 1)^n} \Rightarrow {\left( {{{z + 1} \over z}} \right)^n} = 1$
${\left( {1 + {1 \over z}} \right)^n} = 1$
$\left( {{{1 + \sqrt 3 i} \over 2}} \right) = 1$, then minimum value of n is 6.
2022
Q107
JEE Mains
Numerical
14 Mar 2026
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 ______________.
Correct Answer: 0
Explanation:
$\because$ ${z^2} + \overline z = 0$
Let $z = x + iy$
$\therefore$ ${x^2} - {y^2} + 2ixy + x - iy = 0$
$({x^2} - {y^2} + x) + i(2xy - y) = 0$
$\therefore$ ${x^2} + {y^2} = 0$ and $(2x - 1)y = 0$
if $x = \, + \,{1 \over 2}$ then $y = \, \pm \,{{\sqrt 3 } \over 2}$
And if $y = 0$ then $x = 0, - 1$
$\therefore$ $z = 0 + 0i, - 1 + 0i,{1 \over 2} + {{\sqrt 3 } \over 2}i,{1 \over 2} - {{\sqrt 3 } \over 2}i$
$\therefore$ $\sum {\left( {{R_e}(z) + m(z)} \right) = 0} $
2022
Q108
JEE Mains
Numerical
14 Mar 2026
Let $S = \{ z \in C:|z - 2| \le 1,\,z(1 + i) + \overline z (1 - i) \le 2\} $. Let $|z - 4i|$ attains minimum and maximum values, respectively, at z1 $\in$ S and z2 $\in$ S. If $5(|{z_1}{|^2} + |{z_2}{|^2}) = \alpha + \beta \sqrt 5 $, where $\alpha$ and $\beta$ are integers, then the value of $\alpha$ + $\beta$ is equal to ___________.
Correct Answer: 26
Explanation:
$S$ represents the shaded region shown in the diagram.
Clearly $z_{1}$ will be the point of intersection of $P A$ and given circle.
$P A: 2 x+y=4$ and given circle has equation $(x-2)^{2}+y^{2}=1$
On solving we get
$z_{1}=\left(2-\frac{1}{\sqrt{5}}\right)+\frac{2}{\sqrt{5}} i \Rightarrow\left|z_{1}\right|^{2}=5-\frac{4}{\sqrt{5}}$
$z_{2}$ will be either $B$ or $C$.
$\because P B=\sqrt{17}$ and $P C=\sqrt{13}$ hence $z_{2}=1$
So $5\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)=30-4 \sqrt{5}$
Clearly $\alpha=30$ and $\beta=-4 \Rightarrow \alpha+\beta=26$
2022
Q109
JEE Mains
Numerical
14 Mar 2026
Sum of squares of modulus of all the complex numbers z satisfying $\overline z = i{z^2} + {z^2} - z$ is equal to ___________.
Correct Answer: 2
Explanation:
Let $z=x+i y$
So $2 x=(1+i)\left(x^{2}-y^{2}+2 x y i\right)$
$\Rightarrow 2 x=x^{2}-y^{2}-2 x y\quad$ ...(i) and
$ x^{2}-y^{2}+2 x y=0\quad\dots(ii) $
From (i) and (ii) we get
$ x=0 \text { or } y=-\frac{1}{2} $
When $x=0$ we get $y=0$
When $y=-\frac{1}{2}$ we get $x^{2}-x-\frac{1}{4}=0$
$\Rightarrow \quad x=\frac{-1 \pm \sqrt{2}}{2}$
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 $\left(\frac{-1-\sqrt{2}}{2}\right)-\frac{1}{2} i$
Sum of squares of modulus
$ \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} $
So $2 x=(1+i)\left(x^{2}-y^{2}+2 x y i\right)$
$\Rightarrow 2 x=x^{2}-y^{2}-2 x y\quad$ ...(i) and
$ x^{2}-y^{2}+2 x y=0\quad\dots(ii) $
From (i) and (ii) we get
$ x=0 \text { or } y=-\frac{1}{2} $
When $x=0$ we get $y=0$
When $y=-\frac{1}{2}$ we get $x^{2}-x-\frac{1}{4}=0$
$\Rightarrow \quad x=\frac{-1 \pm \sqrt{2}}{2}$
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 $\left(\frac{-1-\sqrt{2}}{2}\right)-\frac{1}{2} i$
Sum of squares of modulus
$ \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} $
2022
Q110
JEE Mains
Numerical
14 Mar 2026
The number of elements in the set {z = a + ib $\in$ C : a, b $\in$ Z and 1 < | z $-$ 3 + 2i | < 4} is __________.
Correct Answer: 40
Explanation:
at line $y=-2$, we have $(5,-2)(6,-2)(1,-2)(0,-2)$ $\Rightarrow 4$ points
at line $y=-1$, we have $(4,-1)(5,-1)(6,-1)(2,-1)$ $(1,-1)(0,-1) \Rightarrow 6$ points
at line $y=0$, we have $(0,0)(1,0)(2,0)(3,0)(4,0)$ $(5,0)(6,0) \Rightarrow 7$ points
at line $y=1$, we have $(1,1),(2,1),(3,1),(4,1),(5,1)$ i.e. 5 points
symmetrically
at line $y=-5$, we have 5 points
at line $y=-4$, we have 7 points
at line $y=-3$, we have 6 points
So Total integral points $=2(5+7+6)+4$
$ =40 $
2022
Q111
JEE Mains
Numerical
14 Mar 2026
If ${z^2} + z + 1 = 0$, $z \in C$, then
$\left| {\sum\limits_{n = 1}^{15} {{{\left( {{z^n} + {{( - 1)}^n}{1 \over {{z^n}}}} \right)}^2}} } \right|$ is equal to _________.
Correct Answer: 2
Explanation:
$\because$ ${z^2} + z + 1 = 0$
$\Rightarrow$ $\omega$ or $\omega$2
$\because$ $\left| {\sum\limits_{n = 1}^{15} {{{\left( {{z^n} + {{( - 1)}^n}{1 \over {{z^n}}}} \right)}^2}} } \right|$
$ = \left| {\sum\limits_{n = 1}^{15} {{z^{2n}} + \sum\limits_{n = 1}^{15} {{z^{ - 2n}} + 2\,.\,\sum\limits_{n = 1}^{15} {{{( - 1)}^n}} } } } \right|$
$ = \left| {0 + 0 - 2} \right|$
$ = 2$
2022
Q112
JEE Mains
Numerical
14 Mar 2026
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 ___________.
Correct Answer: 80
Explanation:
Here $|z - 3| < 1$
$ \Rightarrow {(x - 3)^2} + {y^2} < 1$
and $z = (4 + 3i) + \overline z (4 - 3i) \le 24$
$ \Rightarrow 4x - 3y \le 12$
$\tan \theta = {4 \over 3}$
$\therefore$ Coordinate of $P = (3 - \cos \theta ,\sin \theta )$
$ = \left( {3 - {3 \over 5},{4 \over 5}} \right)$
$\therefore$ $\alpha + i\beta = {{12} \over 5} + {4 \over 5}i$
$\therefore$ $25(\alpha + \beta ) = 80$
2021
Q113
JEE Mains
MCQ
14 Mar 2026
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 :
A.
$2\sqrt 2 - 1$
B.
$3\sqrt 2 $
C.
$6\sqrt 2 $
D.
$2\sqrt 2 $
2021
Q114
JEE Mains
MCQ
14 Mar 2026
If $S = \left\{ {z \in C:{{z - i} \over {z + 2i}} \in R} \right\}$, then :
A.
S contains exactly two elements
B.
S contains only one element
C.
S is a circle in the complex plane
D.
S is a straight line in the complex plane
2021
Q115
JEE Mains
MCQ
14 Mar 2026
If ${\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq)$, then p and q are roots of the equation :
A.
${x^2} - \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$
B.
${x^2} + \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0$
C.
${x^2} + \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$
D.
${x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0$
2021
Q116
JEE Mains
MCQ
14 Mar 2026
The equation $\arg \left( {{{z - 1} \over {z + 1}}} \right) = {\pi \over 4}$ represents a circle with :
A.
centre at (0, $-$1) and radius $\sqrt 2 $
B.
centre at (0, 1) and radius $\sqrt 2 $
C.
centre (0, 0) and radius $\sqrt 2 $
D.
centre at (0, 1) and radius 2
2021
Q117
JEE Mains
MCQ
14 Mar 2026
Let C be the set of all complex numbers. Let
S1 = {z$\in$C : |z $-$ 2| $\le$ 1} and
S2 = {z$\in$C : z(1 + i) + $\overline z $(1 $-$ i) $\ge$ 4}.
Then, the maximum value of ${\left| {z - {5 \over 2}} \right|^2}$ for z$\in$S1 $\cap$ S2 is equal to :
S1 = {z$\in$C : |z $-$ 2| $\le$ 1} and
S2 = {z$\in$C : z(1 + i) + $\overline z $(1 $-$ i) $\ge$ 4}.
Then, the maximum value of ${\left| {z - {5 \over 2}} \right|^2}$ for z$\in$S1 $\cap$ S2 is equal to :
A.
${{3 + 2\sqrt 2 } \over 4}$
B.
${{5 + 2\sqrt 2 } \over 2}$
C.
${{3 + 2\sqrt 2 } \over 2}$
D.
${{5 + 2\sqrt 2 } \over 4}$
2021
Q118
JEE Mains
MCQ
14 Mar 2026
Let C be the set of all complex numbers. Let
${S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\} $
${S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\} $ and
${S_3} = \{ z \in C||z - \overline z | \ge 8\} $.
Then the number of elements in ${S_1} \cap {S_2} \cap {S_3}$ is equal to :
${S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\} $
${S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\} $ and
${S_3} = \{ z \in C||z - \overline z | \ge 8\} $.
Then the number of elements in ${S_1} \cap {S_2} \cap {S_3}$ is equal to :
A.
1
B.
0
C.
2
D.
Infinite
2021
Q119
JEE Mains
MCQ
14 Mar 2026
Let n denote the number of solutions of the equation z2 + 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 :
A.
1
B.
${4 \over 3}$
C.
${3 \over 2}$
D.
2
2021
Q120
JEE Mains
MCQ
14 Mar 2026
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 :
(Here arg(z) denotes the principal argument of complex number z)
(Here arg(z) denotes the principal argument of complex number z)
A.
${\pi \over 4}$
B.
$ - {{3\pi } \over 4}$
C.
$ - {\pi \over 4}$
D.
${{3\pi } \over 4}$
2021
Q121
JEE Mains
MCQ
14 Mar 2026
Let a complex number be w = 1 $-$ ${\sqrt 3 }$i. Let another complex number z be such that |zw| = 1 and arg(z) $-$ arg(w) = ${\pi \over 2}$. Then the area of the triangle with vertices origin, z and w is equal to :
A.
4
B.
${1 \over 4}$
C.
2
D.
${1 \over 2}$
2021
Q122
JEE Mains
MCQ
14 Mar 2026
If the equation $a|z{|^2} + \overline {\overline \alpha z + \alpha \overline z } + d = 0$ represents a circle where a, d are real constants then which of the following condition is correct?
A.
|$\alpha$|2 $-$ ad $\ne$ 0
B.
|$\alpha$|2 $-$ ad > 0 and a$\in$R $-$ {0}
C.
|$\alpha$|2 $-$ ad $ \ge $ 0 and a$\in$R
D.
$\alpha$ = 0, a, d$\in$R+
2021
Q123
JEE Mains
MCQ
14 Mar 2026
Let S1, S2 and S3 be three sets defined as
S1 = {z$\in$C : |z $-$ 1| $ \le $ $\sqrt 2 $}
S2 = {z$\in$C : Re((1 $-$ i)z) $ \ge $ 1}
S3 = {z$\in$C : Im(z) $ \le $ 1}
Then the set S1 $\cap$ S2 $\cap$ S3 :
S1 = {z$\in$C : |z $-$ 1| $ \le $ $\sqrt 2 $}
S2 = {z$\in$C : Re((1 $-$ i)z) $ \ge $ 1}
S3 = {z$\in$C : Im(z) $ \le $ 1}
Then the set S1 $\cap$ S2 $\cap$ S3 :
A.
has exactly three elements
B.
is a singleton
C.
has infinitely many elements
D.
has exactly two elements
2021
Q124
JEE Mains
MCQ
14 Mar 2026
The area of the triangle with vertices A(z), B(iz) and C(z + iz) is :
A.
1
B.
${1 \over 2}$| z |2
C.
${1 \over 2}$| z + iz |2
D.
${1 \over 2}$
2021
Q125
JEE Mains
MCQ
14 Mar 2026
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 :
A.
8
B.
3
C.
2
D.
$\sqrt 5 $
2021
Q126
JEE Mains
MCQ
14 Mar 2026
Let a complex number z, |z| $\ne$ 1,
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 ____________.
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 ____________.
A.
5
B.
8
C.
6
D.
7
2021
Q127
JEE Mains
MCQ
14 Mar 2026
If $\alpha$, $\beta$ $\in$ R are such that 1 $-$ 2i (here i2 = $-$1) is a root of z2 + $\alpha$z + $\beta$ = 0, then ($\alpha$ $-$ $\beta$) is equal to :
A.
$-$7
B.
7
C.
3
D.
$-$3
2021
Q128
JEE Mains
MCQ
14 Mar 2026
Let the lines (2 $-$ i)z = (2 + i)$\overline z $ and (2 $+$ i)z + (i $-$ 2)$\overline z $ $-$ 4i = 0, (here i2 = $-$1) be normal to a circle C. If the line iz + $\overline z $ + 1 + i = 0 is tangent to this circle C, then its radius is :
A.
${3 \over {2\sqrt 2 }}$
B.
$3\sqrt 2 $
C.
${1 \over {2\sqrt 2 }}$
D.
${3 \over {\sqrt 2 }}$
2021
Q129
JEE Mains
Numerical
14 Mar 2026
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 ______________.
Correct Answer: 5
Explanation:
| z $-$ 2 $-$ 2i | $\le$ 1
| x + iy $-$ 2 $-$ 2i | $\le$ 1
|(x $-$ 2) + i(y $-$ 2)| $\le$ 1
(x $-$ 2)2 + (y $-$ 2)2 $\le$ 1
| 3iz + 6 |max at a + ib
$\left| {3i} \right|\left| {z + {6 \over {3i}}} \right|$
$3{\left| {z - 2i} \right|_{\max }}$
From figure maximum distance at 3 + 2i
a + ib = 3 + 2i = a + b = 3 + 2 = 5 Ans.
| x + iy $-$ 2 $-$ 2i | $\le$ 1
|(x $-$ 2) + i(y $-$ 2)| $\le$ 1
(x $-$ 2)2 + (y $-$ 2)2 $\le$ 1
| 3iz + 6 |max at a + ib
$\left| {3i} \right|\left| {z + {6 \over {3i}}} \right|$
$3{\left| {z - 2i} \right|_{\max }}$
From figure maximum distance at 3 + 2i
a + ib = 3 + 2i = a + b = 3 + 2 = 5 Ans.
2021
Q130
JEE Mains
Numerical
14 Mar 2026
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 _______________.
Correct Answer: 98
Explanation:
Let $z = x + iy$
$\arg \left( {{{x - 2 + iy} \over {x + 2 + iy}}} \right) = {\pi \over 4}$
$\arg (x - 2 + iy) - \arg (x + 2 + iy) = {\pi \over 4}$
${\tan ^{ - 1}}\left( {{y \over {x - 2}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 2}}} \right) = {\pi \over 4}$
${{{y \over {x - 2}} - {y \over {x + 2}}} \over {1 + \left( {{y \over {x - 2}}} \right).\left( {{y \over {x + 2}}} \right)}} = \tan {\pi \over 4} = 1$
${{xy + 2y - xy + 2y} \over {{x^2} - 4 + {y^2}}} = 1$
$4y = {x^2} - 4 + {y^2}$
${x^2} + {y^2} - 4y - 4 = 0$
locus is a circle with center (0, 2) & radius = $2\sqrt 2 $
min. value = ${(AP)^2} = {(OP - OA)^2}$
$ = {\left( {9\sqrt 2 - 2\sqrt 2 } \right)^2}$
$ = {\left( {7\sqrt 2 } \right)^2} = 98$
$\arg \left( {{{x - 2 + iy} \over {x + 2 + iy}}} \right) = {\pi \over 4}$
$\arg (x - 2 + iy) - \arg (x + 2 + iy) = {\pi \over 4}$
${\tan ^{ - 1}}\left( {{y \over {x - 2}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 2}}} \right) = {\pi \over 4}$
${{{y \over {x - 2}} - {y \over {x + 2}}} \over {1 + \left( {{y \over {x - 2}}} \right).\left( {{y \over {x + 2}}} \right)}} = \tan {\pi \over 4} = 1$
${{xy + 2y - xy + 2y} \over {{x^2} - 4 + {y^2}}} = 1$
$4y = {x^2} - 4 + {y^2}$
${x^2} + {y^2} - 4y - 4 = 0$
locus is a circle with center (0, 2) & radius = $2\sqrt 2 $
min. value = ${(AP)^2} = {(OP - OA)^2}$
$ = {\left( {9\sqrt 2 - 2\sqrt 2 } \right)^2}$
$ = {\left( {7\sqrt 2 } \right)^2} = 98$
2021
Q131
JEE Mains
Numerical
14 Mar 2026
Let z1 and z2 be two complex numbers such that $\arg ({z_1} - {z_2}) = {\pi \over 4}$ and z1, z2 satisfy the equation | z $-$ 3 | = Re(z). Then the imaginary part of z1 + z2 is equal to ___________.
Correct Answer: 6
Explanation:
Let z1 = x1 + iy ; z2 = x2 + iy2
z1 $-$ z2 = (x1 $-$ x2) + i(y1 $-$ y2)
$\therefore$ $\arg ({z_1} - {z_2}) = {\pi \over 4}$ $\Rightarrow$ ${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$
${y_1} - {y_2} = {x_1} - {x_2}$ ....... (1)
$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$ .... (2)
$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$ .... (3)
sub (2) & (3)
${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$
$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$
$ = ({x_1} - {x_2})({x_1} + {x_2})$
${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$
z1 $-$ z2 = (x1 $-$ x2) + i(y1 $-$ y2)
$\therefore$ $\arg ({z_1} - {z_2}) = {\pi \over 4}$ $\Rightarrow$ ${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$
${y_1} - {y_2} = {x_1} - {x_2}$ ....... (1)
$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$ .... (2)
$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$ .... (3)
sub (2) & (3)
${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$
$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$
$ = ({x_1} - {x_2})({x_1} + {x_2})$
${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$
2021
Q132
JEE Mains
Numerical
14 Mar 2026
The least positive integer n such that ${{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}},i = \sqrt { - 1} $ is a positive integer, is ___________.
Correct Answer: 6
Explanation:
${{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}} = {{{{(2i)}^n}} \over {{{( - 2i)}^{{{n - 2} \over 2}}}}}$
$ = {{{{(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}}}}}$
This is positive integer for n = 6
$ = {{{{(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}}}}}$
This is positive integer for n = 6
2021
Q133
JEE Mains
Numerical
14 Mar 2026
Let $z = {{1 - i\sqrt 3 } \over 2}$, $i = \sqrt { - 1} $. Then the value of $21 + {\left( {z + {1 \over z}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^3} + .... + {\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$ is ______________.
Correct Answer: 13
Explanation:
$z = {{1 - \sqrt {3i} } \over 2} = {e^{ - i{\pi \over 3}}}$
${z^r} + {1 \over {{z^r}}} = 2\cos \left( { - {\pi \over 3}} \right)r = 2\cos {{r\pi } \over 3}$
$ \Rightarrow 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3} = 8\left( {{{\cos }^3}{{r\pi } \over 3}} \right) = 2\left( {\cos r\pi + 3\cos {{r\pi } \over 3}} \right)} $
$ \Rightarrow 21 + {\left( {z + {1 \over 2}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + ....{\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$
$ = 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3}} $
$ = 21 + \sum\limits_{r = 1}^{21} {\left( {2\cos r\pi + 6\cos {{r\pi } \over 3}} \right)} $
$ = 21 - 2 - 6$
$ = 13$
${z^r} + {1 \over {{z^r}}} = 2\cos \left( { - {\pi \over 3}} \right)r = 2\cos {{r\pi } \over 3}$
$ \Rightarrow 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3} = 8\left( {{{\cos }^3}{{r\pi } \over 3}} \right) = 2\left( {\cos r\pi + 3\cos {{r\pi } \over 3}} \right)} $
$ \Rightarrow 21 + {\left( {z + {1 \over 2}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + ....{\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$
$ = 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3}} $
$ = 21 + \sum\limits_{r = 1}^{21} {\left( {2\cos r\pi + 6\cos {{r\pi } \over 3}} \right)} $
$ = 21 - 2 - 6$
$ = 13$
2021
Q134
JEE Mains
Numerical
14 Mar 2026
If the real part of the complex number $z = {{3 + 2i\cos \theta } \over {1 - 3i\cos \theta }},\theta \in \left( {0,{\pi \over 2}} \right)$ is zero, then the value of sin23$\theta$ + cos2$\theta$ is equal to _______________.
Correct Answer: 1
Explanation:
Re $(z) = {{3 - 6{{\cos }^2}\theta } \over {1 + 9{{\cos }^2}\theta }} = 0$
$\Rightarrow$ $\theta$ = ${{\pi \over 4}}$
Hence, sin23$\theta$ + cos2$\theta$ = 1.
$\Rightarrow$ $\theta$ = ${{\pi \over 4}}$
Hence, sin23$\theta$ + cos2$\theta$ = 1.
2021
Q135
JEE Mains
Numerical
14 Mar 2026
The equation of a circle is Re(z2) + 2(Im(z))2 + 2Re(z) = 0, where z = x + iy. A line which passes through the center of the given circle and the vertex of the parabola, x2 $-$ 6x $-$ y + 13 = 0, has y-intercept equal to ______________.
Correct Answer: 1
Explanation:
Equation of circle is (x2 $-$ y2) + 2y2 + 2x = 0
x2 + y2 + 2x = 0
Centre : ($-$1, 0)
Parabola : x2 $-$ 6x $-$ y + 13 = 0
(x $-$ 3)2 = y $-$ 4
Vertex : (3, 4)
Equation of line $ \equiv y - 0 = {{4 - 0} \over {3 + 1}}(x + 1)$
$y = x + 1$
y-intercept = 1
x2 + y2 + 2x = 0
Centre : ($-$1, 0)
Parabola : x2 $-$ 6x $-$ y + 13 = 0
(x $-$ 3)2 = y $-$ 4
Vertex : (3, 4)
Equation of line $ \equiv y - 0 = {{4 - 0} \over {3 + 1}}(x + 1)$
$y = x + 1$
y-intercept = 1
2021
Q136
JEE Mains
Numerical
14 Mar 2026
Let $S = \left\{ {n \in N\left| {{{\left( {\matrix{
0 & i \cr
1 & 0 \cr
} } \right)}^n}\left( {\matrix{
a & b \cr
c & d \cr
} } \right) = \left( {\matrix{
a & b \cr
c & d \cr
} } \right)\forall a,b,c,d \in R} \right.} \right\}$, where i = $\sqrt { - 1} $. Then the number of 2-digit numbers in the set S is _____________.
Correct Answer: 11
Explanation:
Let $X = \left( {\matrix{
a & b \cr
c & d \cr
} } \right)$ & $A = {\left( {\matrix{
0 & i \cr
1 & 0 \cr
} } \right)^n}$
$\Rightarrow$ AX = IX
$\Rightarrow$ A = I
$ \Rightarrow {\left( {\matrix{ 0 & i \cr 1 & 0 \cr } } \right)^n} = I$
$ \Rightarrow {A^8} = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$
$\Rightarrow$ n is multiple of 8
So, number of 2 digit numbers in the set
S = 11 (16, 24, 32, .........., 96)
$\Rightarrow$ AX = IX
$\Rightarrow$ A = I
$ \Rightarrow {\left( {\matrix{ 0 & i \cr 1 & 0 \cr } } \right)^n} = I$
$ \Rightarrow {A^8} = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$
$\Rightarrow$ n is multiple of 8
So, number of 2 digit numbers in the set
S = 11 (16, 24, 32, .........., 96)
2021
Q137
JEE Mains
Numerical
14 Mar 2026
Let z1, z2 be the roots of the equation z2 + az + 12 = 0 and z1, z2 form an equilateral triangle with origin. Then, the value of |a| is :
Correct Answer: 6
Explanation:
For equilateral triangle with vertices z1, z2 and z3,
$z_1^2 + z_2^2 + z_3^3 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}$
Here one vertex z3 is 0
$ \therefore $ $z_1^2 + z_2^2 = {z_1}{z_2} + 0 + 0$
Given, z1, z2 are roots of ${z^2} + az + 12 = 0$
$ \therefore $ ${z_1} + {z_2} = - a$
${z_1}{z_2} = 12$
$ \therefore $ $z_1^2 + z_2^2 + 2{z_1}{z_2} = {z_1}{z_2} + 2{z_1}{z_2}$
$ \Rightarrow {({z_1} + {z_2})^2} = 3{z_1}{z_2}$
$ \Rightarrow {( - a)^2} = 3 \times 12$
$ \Rightarrow {a^2} = 36$
$ \Rightarrow a = \pm 6$
$ \Rightarrow |a|\, = 6$
$z_1^2 + z_2^2 + z_3^3 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}$
Here one vertex z3 is 0
$ \therefore $ $z_1^2 + z_2^2 = {z_1}{z_2} + 0 + 0$
Given, z1, z2 are roots of ${z^2} + az + 12 = 0$
$ \therefore $ ${z_1} + {z_2} = - a$
${z_1}{z_2} = 12$
$ \therefore $ $z_1^2 + z_2^2 + 2{z_1}{z_2} = {z_1}{z_2} + 2{z_1}{z_2}$
$ \Rightarrow {({z_1} + {z_2})^2} = 3{z_1}{z_2}$
$ \Rightarrow {( - a)^2} = 3 \times 12$
$ \Rightarrow {a^2} = 36$
$ \Rightarrow a = \pm 6$
$ \Rightarrow |a|\, = 6$
2021
Q138
JEE Mains
Numerical
14 Mar 2026
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 $\omega$n is real, is equal to ______________.
Correct Answer: 4
Explanation:
Let z = x + iy
| z + i | = | z $-$ 3i |
$ \Rightarrow $ y = 1
Now
$\omega$ = x2 + y2 $-$ 2x $-$ 2iy + 2
$\omega$ = x2 + 1 $-$ 2x $-$ 2i + 2
Re($\omega$) = x2 $-$ 2x + 3
Re($\omega$) = (x $-$ 1)2 + 2
Re($\omega$)min at x = 1 $ \Rightarrow $ z = 1 + i
Now,
$\omega$ = 1 + 1 $-$ 2 $-$ 2i + 2
$\omega$ = 2(1 $-$ i) = 2$\sqrt 2 {e^{i\left( {{{ - \pi } \over 4}} \right)}}$
$\omega$n = 2$\sqrt 2 {e^{i\left( {{{ - n\pi } \over 4}} \right)}}$
If $\omega$n is real $ \Rightarrow $ n = 4
| z + i | = | z $-$ 3i |
$ \Rightarrow $ y = 1
Now
$\omega$ = x2 + y2 $-$ 2x $-$ 2iy + 2
$\omega$ = x2 + 1 $-$ 2x $-$ 2i + 2
Re($\omega$) = x2 $-$ 2x + 3
Re($\omega$) = (x $-$ 1)2 + 2
Re($\omega$)min at x = 1 $ \Rightarrow $ z = 1 + i
Now,
$\omega$ = 1 + 1 $-$ 2 $-$ 2i + 2
$\omega$ = 2(1 $-$ i) = 2$\sqrt 2 {e^{i\left( {{{ - \pi } \over 4}} \right)}}$
$\omega$n = 2$\sqrt 2 {e^{i\left( {{{ - n\pi } \over 4}} \right)}}$
If $\omega$n is real $ \Rightarrow $ n = 4
2021
Q139
JEE Mains
Numerical
14 Mar 2026
Let z be those complex numbers which satisfy
| z + 5 | $ \le $ 4 and z(1 + i) + $\overline z $(1 $-$ i) $ \ge $ $-$10, i = $\sqrt { - 1} $.
If the maximum value of | z + 1 |2 is $\alpha$ + $\beta$$\sqrt 2 $, then the value of ($\alpha$ + $\beta$) is ____________.
| z + 5 | $ \le $ 4 and z(1 + i) + $\overline z $(1 $-$ i) $ \ge $ $-$10, i = $\sqrt { - 1} $.
If the maximum value of | z + 1 |2 is $\alpha$ + $\beta$$\sqrt 2 $, then the value of ($\alpha$ + $\beta$) is ____________.
Correct Answer: 48
Explanation:
Let, z = x + iy
Given, z(1 + i) + $\overline z $ (1 $-$ i) $ \ge $ $-$ 10
$ \Rightarrow $ z + $\overline z $ + i (z $-$ $\overline z $) $ \ge $ $-$ 10
$ \Rightarrow $ 2x + i (2iy) $ \ge $ $-$ 10
$ \Rightarrow $ x + i2 y $ \ge $ $-$ 5
$ \Rightarrow $ x $-$ y $ \ge $ $-$ 5 ...... (1)
Also given, | z + 5 | $ \le $ 4
$ \Rightarrow $ | z $-$ ($-$5 + 0i) | $ \le $ 4 ...... (2)
It represents a circle whose center at ($-$ 5, 0) and radius 4. z is inside of the circle.
From (1) and (2) z is the shaded region of the diagram.
Now, | z + 1 | = | z $-$ ($-$1 + 0 i) | = distance of z from ($-$1, 0).
Clearly 'p' is the required position of 'z' when | z + 1 | is maximum.
$ \therefore $ P $ \equiv $ ($-$5 $-$ 4 cos45$^\circ$, 0 $-$ 4sin45$^\circ$) = ($-$5$-$2$\sqrt 2 $, $-$2$\sqrt 2 $)
$ \therefore $ (PQ)2|max = 32 + 16$\sqrt 2 $
$ \Rightarrow $ $\alpha$ = 32
$ \Rightarrow $ $\beta$ = 16
Thus, $\alpha$ + $\beta$ = 48
Given, z(1 + i) + $\overline z $ (1 $-$ i) $ \ge $ $-$ 10
$ \Rightarrow $ z + $\overline z $ + i (z $-$ $\overline z $) $ \ge $ $-$ 10
$ \Rightarrow $ 2x + i (2iy) $ \ge $ $-$ 10
$ \Rightarrow $ x + i2 y $ \ge $ $-$ 5
$ \Rightarrow $ x $-$ y $ \ge $ $-$ 5 ...... (1)
Also given, | z + 5 | $ \le $ 4
$ \Rightarrow $ | z $-$ ($-$5 + 0i) | $ \le $ 4 ...... (2)
It represents a circle whose center at ($-$ 5, 0) and radius 4. z is inside of the circle.
From (1) and (2) z is the shaded region of the diagram.
Now, | z + 1 | = | z $-$ ($-$1 + 0 i) | = distance of z from ($-$1, 0).
Clearly 'p' is the required position of 'z' when | z + 1 | is maximum.
$ \therefore $ P $ \equiv $ ($-$5 $-$ 4 cos45$^\circ$, 0 $-$ 4sin45$^\circ$) = ($-$5$-$2$\sqrt 2 $, $-$2$\sqrt 2 $)
$ \therefore $ (PQ)2|max = 32 + 16$\sqrt 2 $
$ \Rightarrow $ $\alpha$ = 32
$ \Rightarrow $ $\beta$ = 16
Thus, $\alpha$ + $\beta$ = 48
2021
Q140
JEE Mains
Numerical
14 Mar 2026
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 _________.
Correct Answer: 310
Explanation:
${(1 + i)^2} = 1 + {i^2} + 2i = 1 - 1 + 2i = 2i$
${(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i$
We know,
$ - {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega $
$ \Rightarrow - 1 + i\sqrt 3 = 2\omega $
and $ - {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2}$
$ \Rightarrow - 1 - i\sqrt 3 = 2{\omega ^2}$
$ \Rightarrow 1 + i\sqrt 3 = - 2{\omega ^2}$
Now, $K = {{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 - i} \right)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 + i} \right)}^{24}}}}$
$ = {{{{(2\omega )}^{21}}} \over {{{\left( {{{(1 - i)}^2}} \right)}^{12}}}} + {{{{( - 2\omega )}^{21}}} \over {{{\left( {{{(1 + i)}^2}} \right)}^{12}}}}$
$ = {{{2^{21}}.{\omega ^{21}}} \over {{{( - 2i)}^{12}}}} + {{{{( - 2)}^{21}}{{({\omega ^2})}^{21}}} \over {{{(2i)}^{12}}}}$ [as ${\omega ^3} = 1$, ${i^4} = 1$]
$ = {{{2^{21}}} \over {{2^{12}}}} - {{{2^{21}}} \over {{2^{12}}}} = 0$
$ \therefore $ $n = \left[ {|K|} \right] = \left[ {|0|} \right] = 0$
Now $\sum\limits_{j = 0}^5 {{{(j + 5)}^2}} - \sum\limits_{j = 0}^5 {(j + 5)} $
= $\sum\limits_{j = 0}^5 {({j^2} + 25 + 10j - j - 5)} $
= $\sum\limits_{j = 0}^5 {({j^2} + 9j + 20)} $
= $\sum\limits_{j = 0}^5 {{j^2}} + 9\sum\limits_{j = 0}^5 {j + 20\sum\limits_{j = 0}^5 1 } $
= ${{5 \times 6 \times 11} \over 6} + 9\left( {{{5 \times 6} \over 2}} \right) + 20 \times 6$
= 55 + 135 + 120
= 310
${(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i$
We know,
$ - {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega $
$ \Rightarrow - 1 + i\sqrt 3 = 2\omega $
and $ - {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2}$
$ \Rightarrow - 1 - i\sqrt 3 = 2{\omega ^2}$
$ \Rightarrow 1 + i\sqrt 3 = - 2{\omega ^2}$
Now, $K = {{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 - i} \right)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 + i} \right)}^{24}}}}$
$ = {{{{(2\omega )}^{21}}} \over {{{\left( {{{(1 - i)}^2}} \right)}^{12}}}} + {{{{( - 2\omega )}^{21}}} \over {{{\left( {{{(1 + i)}^2}} \right)}^{12}}}}$
$ = {{{2^{21}}.{\omega ^{21}}} \over {{{( - 2i)}^{12}}}} + {{{{( - 2)}^{21}}{{({\omega ^2})}^{21}}} \over {{{(2i)}^{12}}}}$ [as ${\omega ^3} = 1$, ${i^4} = 1$]
$ = {{{2^{21}}} \over {{2^{12}}}} - {{{2^{21}}} \over {{2^{12}}}} = 0$
$ \therefore $ $n = \left[ {|K|} \right] = \left[ {|0|} \right] = 0$
Now $\sum\limits_{j = 0}^5 {{{(j + 5)}^2}} - \sum\limits_{j = 0}^5 {(j + 5)} $
= $\sum\limits_{j = 0}^5 {({j^2} + 25 + 10j - j - 5)} $
= $\sum\limits_{j = 0}^5 {({j^2} + 9j + 20)} $
= $\sum\limits_{j = 0}^5 {{j^2}} + 9\sum\limits_{j = 0}^5 {j + 20\sum\limits_{j = 0}^5 1 } $
= ${{5 \times 6 \times 11} \over 6} + 9\left( {{{5 \times 6} \over 2}} \right) + 20 \times 6$
= 55 + 135 + 120
= 310
2021
Q141
JEE Mains
Numerical
14 Mar 2026
If the least and the largest real values of a, for which the
equation z + $\alpha $|z – 1| + 2i = 0 (z $ \in $ C and i = $\sqrt { - 1} $) has a solution, are p and q respectively; then 4(p2 + q2) is equal to __________.
equation z + $\alpha $|z – 1| + 2i = 0 (z $ \in $ C and i = $\sqrt { - 1} $) has a solution, are p and q respectively; then 4(p2 + q2) is equal to __________.
Correct Answer: 10
Explanation:
$x + iy + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} + 2i = 0$
$ \therefore $ y + 2 = 0 and $x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0$
y = $-$2 & ${x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4)$
${\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^2} + 5{\alpha ^2} = 0$
$x \in R \Rightarrow D \ge 0$
$4{\alpha ^4} - 4({\alpha ^2} - 1)5{\alpha ^2} \ge 0$
${\alpha ^2}[4{\alpha ^2} - 2{\alpha ^2} + 20] \ge 0$
${\alpha ^2}[ - 16{\alpha ^2} + 20] \ge 0$
${\alpha ^2}\left[ {{\alpha ^2} - {5 \over 4}} \right] \le 0$
$0 \le {\alpha ^2} \le {5 \over 4}$
$ \therefore $ ${\alpha ^2} \in \left[ {0,{5 \over 4}} \right]$
$ \therefore $ $\alpha \in \left[ { - {{\sqrt 5 } \over 2},{{\sqrt 5 } \over 2}} \right]$
then $4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10$
$ \therefore $ y + 2 = 0 and $x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0$
y = $-$2 & ${x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4)$
${\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^2} + 5{\alpha ^2} = 0$
$x \in R \Rightarrow D \ge 0$
$4{\alpha ^4} - 4({\alpha ^2} - 1)5{\alpha ^2} \ge 0$
${\alpha ^2}[4{\alpha ^2} - 2{\alpha ^2} + 20] \ge 0$
${\alpha ^2}[ - 16{\alpha ^2} + 20] \ge 0$
${\alpha ^2}\left[ {{\alpha ^2} - {5 \over 4}} \right] \le 0$
$0 \le {\alpha ^2} \le {5 \over 4}$
$ \therefore $ ${\alpha ^2} \in \left[ {0,{5 \over 4}} \right]$
$ \therefore $ $\alpha \in \left[ { - {{\sqrt 5 } \over 2},{{\sqrt 5 } \over 2}} \right]$
then $4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10$
2020
Q142
JEE Mains
MCQ
14 Mar 2026
Let z = x + iy be a non-zero complex number
such that ${z^2} = i{\left| z \right|^2}$, where i = $\sqrt { - 1} $ , then z lies
on the :
A.
line, y = –x
B.
real axis
C.
line, y = x
D.
imaginary axis
2020
Q143
JEE Mains
MCQ
14 Mar 2026
The region represented by
{z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1} is also given by the
inequality : {z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1}
{z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1} is also given by the
inequality : {z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1}
A.
y2 $ \le $ $2\left( {x + {1 \over 2}} \right)$
B.
y2 $ \le $ ${x + {1 \over 2}}$
C.
y2 $ \ge $ 2(x + 1)
D.
y2 $ \ge $ x + 1
2020
Q144
JEE Mains
MCQ
14 Mar 2026
The value of ${\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}}$ is :
A.
–215i
B.
–215
C.
215i
D.
65
2020
Q145
JEE Mains
MCQ
14 Mar 2026
If the four complex numbers $z,\overline z ,\overline z - 2{\mathop{\rm Re}\nolimits} \left( {\overline z } \right)$ and $z-2Re(z)$ represent the vertices of a square of
side 4 units in the Argand plane, then $|z|$ is equal to :
A.
4$\sqrt 2 $
B.
4
C.
2
D.
2$\sqrt 2 $
Let $z = x + iy$
2020
Q146
JEE Mains
MCQ
14 Mar 2026
If a and b are real numbers such that
${\left( {2 + \alpha } \right)^4} = a + b\alpha $
where $\alpha = {{ - 1 + i\sqrt 3 } \over 2}$ then a + b is equal to :
${\left( {2 + \alpha } \right)^4} = a + b\alpha $
where $\alpha = {{ - 1 + i\sqrt 3 } \over 2}$ then a + b is equal to :
A.
33
B.
9
C.
24
D.
57
2020
Q147
JEE Mains
MCQ
14 Mar 2026
Let $u = {{2z + i} \over {z - ki}}$, z = x + iy and k > 0. If the curve represented
by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is :
by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is :
A.
2
B.
4
C.
1/2
D.
3/2
2020
Q148
JEE Mains
MCQ
14 Mar 2026
If z1
, z2
are complex numbers such that
Re(z1) = |z1 – 1|, Re(z2) = |z2 – 1| , and
arg(z1 - z2) = ${\pi \over 6}$, then Im(z1 + z2 ) is equal to :
Re(z1) = |z1 – 1|, Re(z2) = |z2 – 1| , and
arg(z1 - z2) = ${\pi \over 6}$, then Im(z1 + z2 ) is equal to :
A.
${{\sqrt 3 } \over 2}$
B.
${1 \over {\sqrt 3 }}$
C.
${2 \over {\sqrt 3 }}$
D.
${2\sqrt 3 }$
2020
Q149
JEE Mains
MCQ
14 Mar 2026
The imaginary part of
${\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}$ can be :
${\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}$ can be :
A.
-2$\sqrt 6 $
B.
6
C.
$\sqrt 6 $
D.
-$\sqrt 6 $
2020
Q150
JEE Mains
MCQ
14 Mar 2026
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 :
${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$ is :
A.
${1 \over 2}\left( {\sqrt 3 - i} \right)$
B.
-${1 \over 2}\left( {\sqrt 3 - i} \right)$
C.
$ - {1 \over 2}\left( {1 - i\sqrt 3 } \right)$
D.
${1 \over 2}\left( {1 - i\sqrt 3 } \right)$