Complex Numbers
523 Questions
Start JEE Mains Test
2022
Q301
BITSAT
MCQ
11 Jun 2026
The smallest positive integral value of n such that ${\left[ {{{1 + \sin {\pi \over 8} + i\cos {\pi \over 8}} \over {1 + \sin {\pi \over 8} - i\cos {\pi \over 8}}}} \right]^n}$ is purely imaginary, is equal to
A.
4
B.
3
C.
2
D.
8
2021
Q302
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
Q303
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
Q304
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
Q305
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
Q306
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
Q307
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
Q308
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
Q309
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
Q310
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
Q311
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
Q312
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
Q313
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
Q314
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
Q315
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
Q316
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
Q317
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
Q318
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
Q319
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
Q320
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
Q321
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
Q322
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
Q323
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
Q324
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
Q325
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
Q326
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
Q327
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
Q328
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
Q329
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
Q330
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 $4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10$
2021
Q331
JEE Advanced
MCQ
14 Mar 2026
Let $\theta_1, \theta_2, \ldots, \theta_{10}$ be positive valued angles (in radian) such that $\theta_1+\theta_2+\cdots+\theta_{10}=2 \pi$. Define the complex numbers $z_1=e^{i \theta_1}, z_k=z_{k-1} e^{i \theta_k}$ for $k=2,3, \ldots, 10$, where $i=\sqrt{-1}$. Consider the statements $P$ and $Q$ given below:
$P:\left| {{z_2} - {z_1}} \right| + \left| {{z_3} - {z_2}} \right| + ..... + \left| {{z_{10}} - {z_9}} \right| + \left| {{z_1} - {z_{10}}} \right| \le 2\pi $
$Q:\left| {z_2^2 - z_1^2} \right| + \left| {z_3^2 - z_2^2} \right| + .... + \left| {z_{10}^2 - z_9^2} \right| + \left| {z_1^2 - z_{10}^2} \right| \le 4\pi $
Then,
$P:\left| {{z_2} - {z_1}} \right| + \left| {{z_3} - {z_2}} \right| + ..... + \left| {{z_{10}} - {z_9}} \right| + \left| {{z_1} - {z_{10}}} \right| \le 2\pi $
$Q:\left| {z_2^2 - z_1^2} \right| + \left| {z_3^2 - z_2^2} \right| + .... + \left| {z_{10}^2 - z_9^2} \right| + \left| {z_1^2 - z_{10}^2} \right| \le 4\pi $
Then,
A.
P is TRUE and Q is FALSE
B.
Q is TRUE and P is FALSE
C.
both P and Q are TRUE
D.
both P and Q are FALSE
2021
Q332
JEE Advanced
MSQ
14 Mar 2026
For any complex number w = c + id, let $\arg (w) \in ( - \pi ,\pi ]$, where $i = \sqrt { - 1} $. Let $\alpha$ and $\beta$ be real numbers such that for all complex numbers z = x + iy satisfying $\arg \left( {{{z + \alpha } \over {z + \beta }}} \right) = {\pi \over 4}$, the ordered pair (x, y) lies on the circle ${x^2} + {y^2} + 5x - 3y + 4 = 0$, Then which of the following statements is (are) TRUE?
A.
$\alpha$ = $-$1
B.
$\alpha$$\beta$ = 4
C.
$\alpha$$\beta$ = $-$4
D.
$\beta$ = 4
2021
Q333
AP-EAPCET
MCQ
20 May 2026
Let $Z_1, Z_2$ and $Z_3$ be three non zero complex numbers such that $a=\left|Z_1\right|, b=\left|Z_2\right|$ and $c=\left|Z_3\right|$, if the determinant $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=0$, then
A.
$\left|Z_1\right|=\left|Z_2\right|=\left|Z_3\right|=a b c$
B.
$\left|Z_1\right|+\left|Z_2\right|+\left|Z_3\right|=0$
C.
$\left|Z_1\right|+\left|Z_2\right|+\left|Z_3\right|=a b c$
D.
$\left|Z_1-Z_2\right|=\left|Z_2-Z_3\right|$
2021
Q334
AP-EAPCET
MCQ
20 May 2026
If $\left|z_1+z_2\right|^2=\left|z_1\right|^2+\left|z_2\right|^2$, where $z_1$ and $z_2$ are two complex numbers, then
A.
$\frac{z_1}{z_2}$ is purely real
B.
$\frac{z_1}{z_2}$ is purely imaginary
C.
$\arg \left(\frac{z_1}{z_2}\right)=\frac{\pi}{4}$
D.
$\left|\frac{z_1}{z_2}\right|=1$
2021
Q335
AP-EAPCET
MCQ
20 May 2026
A real value of $x$ will satisfy the equation, $\left(\frac{3-4 i x}{3+4 i x}\right)=\alpha-i \beta,(\alpha, \beta$ are real $)$, if
A.
$\alpha^2-\beta^2=-1$
B.
$\alpha^2-\beta^2=1$
C.
$\alpha^2+\beta^2=1$
D.
$\alpha^2-\beta^2=2$
2021
Q336
AP-EAPCET
MCQ
20 May 2026
What is the value of $(1-i \sqrt{3})^9$ is equal to
A.
$2^9$
B.
$-2^9$
C.
$2^9 i$
D.
$-2^9 i$
2021
Q337
AP-EAPCET
MCQ
20 May 2026
$\left(\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4} i\right)^{2020}$ is equal to
A.
$\frac{1}{2}+\frac{\sqrt{3}}{2} i$
B.
$\frac{-1}{2}+\frac{\sqrt{3}}{2} i$
C.
$\frac{-1}{2}-\frac{\sqrt{3}}{2} i$
D.
$\frac{1}{2}-\frac{\sqrt{3}}{2} i$
2021
Q338
AP-EAPCET
MCQ
20 May 2026
If $z_1=2+3 i$ and $z_2=3+2 i$, where $i=\sqrt{-1}$, then $\left[\begin{array}{cc}z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1\end{array}\right]\left[\begin{array}{cc}\bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1\end{array}\right]$ is equal to
A.
$13 I$
B.
$I$
C.
$26 I$
D.
Zero matrix
2021
Q339
AP-EAPCET
MCQ
20 May 2026
The radius of the circle represented by $(1+i)(1+3i)(1+7i)=x+iy$ is $(i=\sqrt{-1})$.
A.
1000
B.
10$\sqrt{10}$
C.
10000
D.
100
2021
Q340
AP-EAPCET
MCQ
20 May 2026
If $1, \alpha_1, \alpha_2, \alpha_3$ and $\alpha_4$ are the roots of $z^5-1=0$ and $\omega$ is a cube root of units, then $(\omega-1)\left(\omega-\alpha_1\right)\left(\omega-\alpha_2\right)\left(\omega-\alpha_3\right)\left(\omega-\alpha_4\right)+\omega$ is equal to
A.
0
B.
$-$1
C.
$-$2
D.
1
2021
Q341
AP-EAPCET
MCQ
20 May 2026
If $a > 0$ and $z=x+i y$, then $\log _{\cos ^2 \theta}|z-a|>\log _{\cos ^2 \theta}|z-a i|,(\theta \in R)$ implies
A.
$x>y$
B.
$x < y$
C.
$x+y=\cos \theta$
D.
$x+y<0$
2021
Q342
AP-EAPCET
MCQ
20 May 2026
If one root of the equation $i x^2-2(i+1) x+(2-i)=0$ is $(2-i)$, then the other root is
A.
$-i$
B.
$2+i$
C.
$i$
D.
$2-i$
2021
Q343
AP-EAPCET
MCQ
20 May 2026
If $|z-2|=|z-1|$, where $z$ is a complex number, then locus $z$ is a straight line
A.
Parallel to $X$ - axis
B.
Parallel to $Y$-axis
C.
Parallel to $y=x$
D.
Parallel to $y=-x$
2021
Q344
AP-EAPCET
MCQ
20 May 2026
If ${\left( {{{1 + i} \over {1 - i}}} \right)^m} = 1$, then m cannot be equal to
A.
1934
B.
2024
C.
2172
D.
10100
2021
Q345
AP-EAPCET
MCQ
20 May 2026
$(\sin \theta-i \cos \theta)^3$ is equal to
A.
$i^3(\cos 3 \theta+i \sin 3 \theta)$
B.
$\cos 3 \theta+i \sin 3 \theta$
C.
$\sin 3 \theta-i \cos 3 \theta$
D.
$(-i)^3(\cos 3 \theta+i \sin 3 \theta)$
2021
Q346
AP-EAPCET
MCQ
20 May 2026
Real part of $(\cos 4+i \sin 4+1)^{2020}$ is
A.
$2^{2020} \cos ^{2020} 2 \cos 2020$
B.
$2^{2020} \cos ^{2020} 2 \cos 4040$
C.
$2^{1020} \cos ^{2020} 2 \cos 4040$
D.
$2^{2020} \cos ^{2020} 1 \cos 2020$
2021
Q347
BITSAT
MCQ
11 Jun 2026
If Re(z + 2) = | z $-$ 2 |, then the locus of z is
A.
parabola
B.
circle
C.
ellipse
D.
hyperbola
2020
Q348
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
Q349
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
Q350
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