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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 337 questions
1 Medium
1 of 337
Application +4 | -1
If ∣z∣=1|z|=1 and z≠±1z \neq \pm 1, then for any complex number z,z1−z2z, \frac{z}{1-z^2}
View answer and explanation
Correct option: D
Lies on Y-axis
Option D (Lies on Y-axis) is correct because z1−z2\frac{z}{1-z^2} simplifies to a purely imaginary number, indicating it lies on the Y-axis.
2 Medium
2 of 337
Application +4 | -1
The roots of the cubic equation (z+ab)3=a3(z+a b)^3=a^3, where a≠0a \neq 0 represents the vertices of a triangle. If α,β\alpha, \beta and γ\gamma are the roots, then find the value of ∣α−β∣|\alpha-\beta|.
View answer and explanation
Correct option: D
3∣ab∣\sqrt{3}|a b|
Option D (3∣ab∣\sqrt{3}|a b|) is correct because the roots of the cubic equation represent vertices of an equilateral triangle, and the distance between two vertices is 3∣ab∣\sqrt{3}|ab|.
3 Medium
3 of 337
Application +4 | -1
Number of solutions of the equation z2+∣z∣2=0,{z^2} + {\left| z \right|^2} = 0, where z∈Cz \in C is
View answer and explanation
Correct option: D
Infinity many
Option D (Infinity many) is correct because the equation simplifies to x=0, allowing y to be any real number, resulting in infinite solutions.
4 Medium
4 of 337
Application +4 | -1
3+2isin⁡θ1−2isin⁡θ  \frac{{3 + 2i\sin \mathrm{\theta }}}{{1 - 2i\sin \mathrm{\theta }}}\; will be real, if θ\theta is
View answer and explanation
Correct option: C
nπn\pi
Option C (nπn\pi) is correct because for the given expression to be real, the imaginary part must be zero, which occurs when sin⁡θ=0\sin \theta = 0, implying θ=nπ\theta = n\pi.
5 Medium
5 of 337
Application +4 | -1
If z=4+i7z=4+i \sqrt{7}, then value of z3−4z2−9z+91z^3-4 z^2-9 z+91 equals
View answer and explanation
Correct option: D
-1
Option D (-1) is correct because after substituting z=4+i7z=4+i \sqrt{7} into the expression and simplifying, the result is -1.
6 Medium
6 of 337
Application +4 | -1
If (x+iy)=1+2i3+4i\left( {x + iy} \right) = \sqrt {\frac{{1 + 2i}}{{3 + 4i}}}, then (x2+y2)2{\left( {{x^2} + {y^2}} \right)^2} is equal to
View answer and explanation
Correct option: B
15\frac{1}{5}
Option B 15\frac{1}{5} is correct because it is the value of (x2+y2)2{\left( {{x^2} + {y^2}} \right)^2} obtained by taking the modulus of both sides of the given equation and simplifying.
7 Medium
7 of 337
Application +4 | -1
If ∣z1∣=∣z2∣\left|z_1\right|=\left|z_2\right| and arg⁡(z1)+arg⁡(z2)=π/2\arg \left(z_1\right)+\arg \left(z_2\right)=\pi / 2, then
View answer and explanation
Correct option: B
z1z2z_1 z_2 is purely imaginary
The correct answer is B (z1z2z_1 z_2 is purely imaginary) because the product of two complex numbers with arguments summing to π/2\pi/2 results in a purely imaginary number.
8 Medium
8 of 337
Application +4 | -1
{1+cos⁡π8+isin⁡π81+cos⁡π8−isin⁡π8}8=\left\{\frac{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}\right\}^8=
View answer and explanation
Correct option: B
1−i1-i
Option D (-1) is correct because simplifying the expression using Euler's formula and De Moivre's theorem leads to -1.
9 Medium
9 of 337
Application +4 | -1
If (cos⁡θ+isin⁡θ)(cos⁡2θ+isin⁡2θ)…(cos⁡nθ+isin⁡nθ)=1(\cos \theta+i \sin \theta)(\cos 2 \theta+i \sin 2 \theta) \ldots(\cos n \theta+i \sin n \theta)=1, then θ=\theta=
View answer and explanation
Correct option: C
4kπn(n+1),k∈Z\frac{4 k \pi}{n(n+1)}, k \in Z
Option C (4kπn(n+1),k∈Z\frac{{4 k \pi}}{{n(n+1)}}, k \in Z) is correct because it is derived by equating the sum of angles to 2kπ2k\pi and solving for θ\theta.
10 Medium
10 of 337
Application +4 | -1
If zz is a complex number such that ∣z∣≠0|z| \neq 0 and Re⁡(z)=0\operatorname{Re}(z)=0, then
View answer and explanation
Correct option: B
Im⁡(z2)=0\operatorname{Im}\left(z^2\right)=0
Option B (Im⁡(z2)=0\operatorname{Im}(z^2)=0) is correct because if the real part of z is 0, then z is purely imaginary, and the square of a purely imaginary number is a real number, meaning its imaginary part is 0.
11 Medium
11 of 337
Analyzing +4 | -1
If 1,ω1, \omega and ω2\omega^2 are the three cube roots of unity and α,β\alpha, \beta and γ\gamma are the roots of p,p<0p, p < 0, then for any x,yx, y and zz, the expression xα+yβ+zγxβ+yγ+zα\frac{x \alpha+y \beta+z \gamma}{x \beta+y \gamma+z \alpha} equals
View answer and explanation
Correct option: C
ω2\omega^2
Option C (ω2\omega^2) is correct because by substituting the cube roots of unity and simplifying the expression, it reduces to ω2\omega^2.
12 Medium
12 of 337
Application +4 | -1
The least positive integer nn for which (1+i1−i)n=2πsin⁡−1(1+x22x),{\left( {\frac{{1 + i}}{{1 - i}}} \right)^n} = \frac{2}{\pi }{\sin ^{ - 1}}\left( {\frac{{1 + {x^2}}}{{2x}}} \right), where x>0,x > 0, is
View answer and explanation
Correct option: B
4
Option B (4) is correct because it is the smallest positive integer nn that satisfies the equation after simplifying the complex number and the inverse sine function.
13 Medium
13 of 337
Analyzing +4 | -1
If A(z1),B(z2),C(z3)A\left(z_1\right), B\left(z_2\right), C\left(z_3\right) are vertices of a triangle such that z3=(z2−iz11−i)z_3=\left(\frac{z_2-i z_1}{1-i}\right) and ∣z1∣=3,∣z2∣=4\left|z_1\right|=3,\left|z_2\right|=4 and ∣z2+iz1∣=∣z1∣+∣z2∣\left|z_2+i z_1\right|=\left|z_1\right|+\left|z_2\right| then area of triangle ABC is
View answer and explanation
Correct option: D
254\frac{25}{4}
Option D (254\frac{25}{4}) is correct because the area of the triangle ABC is calculated using the collinearity of points and geometric properties of complex numbers.
14 Medium
14 of 337
Application +4 | -1
If ∣z∣<1/2|z|<1 / 2 and 4∣(1+i)z3+iz∣<k4\left|(1+i) z^3+i z\right| < k, then the least integral value of kk is
View answer and explanation
Correct option: C
3
Option C (3) is correct because using the triangle inequality and the given condition ∣z∣<1/2|z| < 1/2, the expression is bounded by 3.
15 Medium
15 of 337
Application +4 | -1
If x+iy=32+cos⁡θ+isin⁡θx+i y=\frac{3}{2+\cos \theta+i \sin \theta}, then x2+y2x^2+y^2 is equal to
View answer and explanation
Correct option: B
4x−34 x-3
The value of p2+q2p^2+q^2 is (a2+1)24a2−1\frac{{\left(a^2+1\right)^2}}{{4 a^2-1}} because it is obtained by multiplying the complex number with its conjugate.

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16 Medium
16 of 337
Application +4 | -1
If z=x+iy,xz=x+i y, x and yy are real, then ∣x∣+∣y∣≤k∣z∣|x|+|y| \leq k|z|, where kk is equal to
View answer and explanation
Correct option: B
2\sqrt{2}
Option B is correct because the inequality ∣x∣+∣y∣≤2∣z∣|x|+|y| \leq \sqrt{2}|z| holds true, where k=2k = \sqrt{2}.
17 Medium
17 of 337
Application +4 | -1
If zz is a complex number, then the minimum value of ∣z∣+∣z−1∣\left| z \right| + \left| {z - 1} \right| is
View answer and explanation
Correct option: A
1
The correct answer is A (1) because the minimum value is achieved when z lies on the line segment connecting 0 and 1.
18 Medium
18 of 337
Application +4 | -1
The amplitude of sin⁡π5+i(1−cos⁡π5)\sin \frac{\pi }{5} + i\left( {1 - \cos \frac{\pi }{5}} \right) is
View answer and explanation
Correct option: C
π10\frac{\pi }{{10}}
Option C (π10\frac{\pi }{{10}}) is correct because it accurately calculates the argument (amplitude) of the given complex number using trigonometric identities.
19 Medium
19 of 337
Application +4 | -1
If zz is any complex number such that z+1z=1z+\frac{1}{z}=1, then the value of z99+1z99z^{99}+\frac{1}{z^{99}} is
View answer and explanation
Correct option: D
-2
Option B (-2) is correct because the given condition implies that z is a complex cube root of unity (specifically −ω-\omega or −ω2-\omega^2), and substituting this into the expression yields -2.
20 Medium
20 of 337
Application +4 | -1
If α=−1+i3\alpha=-1+i \sqrt{3} and nn is a positive integer which is not a multiple of 3 , then
α2n+2nαn+22n=\alpha^{2 n}+2^n \alpha^n+2^{2 n}=
View answer and explanation
Correct option: C
0
Option C (0) is correct because when α is expressed in terms of ω, the expression simplifies to 0 since n is not a multiple of 3.
21 Medium
21 of 337
Application +4 | -1
If the cube root of unity are 1,  ω,  ω21,\;\omega ,\;{\omega ^2} then the roots of the equation (x−1)3+8=0{\left( {x - 1} \right)^3} + 8 = 0, are
View answer and explanation
Correct option: B
−1,  1−2ω,  1−2ω2- 1,\;1 - 2\omega ,\;1 - 2{\omega ^2}
Option B (−1,  1−2ω,  1−2ω2- 1,\;1 - 2\omega ,\;1 - 2{\omega ^2}) is correct because it represents the roots of the given equation after simplification and using the properties of cube roots of unity.
22 Medium
22 of 337
Understanding +4 | -1
∣z1+z2∣2+∣z1−z2∣2\left|z_{1}+z_{2}\right|^{2}+\left|z_{1}-z_{2}\right|^{2} is equal to
View answer and explanation
Correct option: B
2(∣z1∣2+∣z2∣2)2\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)
Option B is correct because |z1+z2|^2 + |z1-z2|^2 simplifies to 2(|z1|^2 + |z2|^2) using the properties of complex numbers.
23 Medium
23 of 337
Analyzing +4 | -1
If z=x+iyz=x+i y and w=1−izz−iw=\frac{1-i z}{z-i}, then ∣w∣=1|w|=1 implies, that, in the complex plane
View answer and explanation
Correct option: B
zz lies on the real axis
Option B (zz lies on the real axis) is correct because ∣w∣=1|w|=1 implies ∣z+i∣=∣z−i∣|z+i|=|z-i|, which means zz is equidistant from ii and −i-i, thus lying on the real axis.
24 Medium
24 of 337
Application +4 | -1
The real part of tan⁡(α+iβ)\tan (\alpha+i \beta) is
View answer and explanation
Correct option: A
2sin⁡2α2cos⁡2α+e2β+e−2β\frac{2 \sin 2 \alpha}{2 \cos 2 \alpha+\mathrm{e}^{2 \beta}+\mathrm{e}^{-2 \beta}}
Option A is correct because the real part of tan⁡(α+iβ)\tan(\alpha+i \beta) is derived using trigonometric identities and Euler's formula.
25 Medium
25 of 337
Application +4 | -1
If ∣z+2iz−i∣=2\left|\frac{z+2 i}{z-i}\right|=2 is a circle, then centre of the circle is
View answer and explanation
Correct option: B
(0,2)(0,2)
Option A (\left[\begin{array}{cc}1 & -2023 i \\ 0 & 1\end{array}\right]) is correct because after simplifying the expression and using the properties of matrix multiplication, the result is a matrix with -2023i in the top right corner.
26 Medium
26 of 337
Application +4 | -1
If ω=cos⁡πn+isin⁡πn\omega=\cos \frac{\pi}{n}+i \sin \frac{\pi}{n}, then value of 1+ω+ω2+…+ωn−11+\omega+\omega^2+\ldots+\omega^{n-1} is
View answer and explanation
Correct option: C
1+icot⁡(π/2n)1+i \cot (\pi / 2 n)
The correct answer is D (π/2\pi / 2) because the odd functions integrate to zero over the symmetric interval, leaving only the integral of the constant function 1.
27 Medium
27 of 337
Application +4 | -1
If ∣z+1∣+∣z−3∣≤10|z+1|+|z-3| \leq 10, then the range of values of ∣z−7∣|z-7| is
View answer and explanation
Correct option: B
[3,13][3,13]
Option B ([3,13]) is correct because it represents the range of distances from a point on the ellipse to the point (7,0).
28 Medium
28 of 337
Application +4 | -1
The value of log⁡35×log⁡2527×log⁡497log⁡813\frac{{{{\log }_3}5 \times {{\log }_{25}}27 \times {{\log }_{49}}7}}{{{{\log }_{81}}3}} is
View answer and explanation
Correct option: D
3
Option D (3) is correct because the expression simplifies to 3 using the change of base formula for logarithms.
29 Medium
29 of 337
Application +4 | -1
If z=1+7i(2−i)2z=\frac{1+7 i}{(2-i)^2}, then the polar form of zz is
View answer and explanation
Correct option: C
2[cos⁡3π4+isin⁡3π4]\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]
Option C (2[cos⁡3π4+isin⁡3π4]\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]) is correct because it represents the polar form of the given complex number after simplification.
30 Medium
30 of 337
Understanding +4 | -1
If z≠0z \neq 0, then ∫x=0100[arg⁡∣z∣]dx\int_{x=0}^{100}[\arg |z|] d x is (where [∗]\left[{ }^*\right] denotes the greatest integer function)
View answer and explanation
Correct option: A
0
Option A (0) is correct because the argument of the absolute value of any complex number is always 0, and the greatest integer function of 0 is 0.
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