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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Mixed difficulty

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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 979 questions
1 Easy
1 of 979
Application +4 | -1
The number of non-zero integer solutions of the equation ∣1−i∣x=2x{\left| {1 - i} \right|^x} = {2^x} is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the equation |1 - i|^x = 2^x has only one solution, x = 0, which is not a non-zero integer.
2 Medium
2 of 979
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.
3 Easy
3 of 979
Application +4 | -1
Let ∣zr−r∣≤r,∀r=1,2,3,…,n\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n, then ∣∑r=1nzr∣\left|\sum\limits_{r=1}^n z_r\right| is less than
View answer and explanation
Correct option: C
n(n+1)n(n+1)
Option C (n(n+1)n(n+1)) is correct because it represents the upper bound of the sum of the complex numbers based on the given inequality.
4 Medium
4 of 979
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|.
5 Medium
5 of 979
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.
6 Easy
6 of 979
Application +4 | -1
If 2x=−1+3i2x = - 1 + \sqrt 3 i, then the value of (1−x2+x)6−(1−x+x2)6{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6} is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because substituting x = ω (cube root of unity) simplifies the expression to 0.
7 Easy
7 of 979
Application +4 | -1
The amplitude of 1+i33+i\frac{{1 + i\sqrt 3 }}{{\sqrt 3 + i}} is
View answer and explanation
Correct option: D
π6\frac{\pi }{6}
The correct answer is D (π6\frac{\pi}{6}) because it is the argument of the simplified complex number.
8 Medium
8 of 979
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.
9 Easy
9 of 979
Analyzing +4 | -1
The smallest positive integral value of nn such that [1+sin⁡π8+icos⁡π81+sin⁡π8−icos⁡π8]n{\left[ {\frac{{1 + \sin \frac{\pi }{8} + i\cos \frac{\pi }{8}}}{{1 + \sin \frac{\pi }{8} - i\cos \frac{\pi }{8}}}} \right]^n} is purely imaginary, is equal to
View answer and explanation
Correct option: A
4
The correct answer is A (4) because when n=4, the expression becomes purely imaginary, satisfying the condition.
10 Easy
10 of 979
Application +4 | -1
If (3+i)(z+zˉ)−(2+i)(z−zˉ)+14i=0\left( {3 + i} \right)\left( {z + \bar z} \right) - \left( {2 + i} \right)\left( {z - \bar z} \right) + 14i = 0, then zzˉz\bar z is equal to
View answer and explanation
Correct option: C
10
Option C (10) is correct because substituting z = x + iy into the given equation and solving the resulting system of equations yields x = -1 and y = 3, thus zzˉ\bar z = 10.
11 Medium
11 of 979
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.
12 Medium
12 of 979
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.
13 Easy
13 of 979
Analyzing +4 | -1
If z1,z2,z3z_1, z_2, z_3 and z4z_4 are the four complex numbers represented by the vertices of a quadrilateral taken in order such that z1−z4=z2−z3z_1-z_4=z_2-z_3 and amp⁡(z4−z1z2−z1)=π2\operatorname{amp}\left(\frac{z_4-z_1}{z_2-z_1}\right)=\frac{\pi}{2}, then the quadrilateral is a
View answer and explanation
Correct option: C
cyclic quadrilateral
Option C (cyclic quadrilateral) is correct because the given conditions imply that the quadrilateral is a rectangle, which is a cyclic quadrilateral.
14 Medium
14 of 979
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.
15 Easy
15 of 979
Application +4 | -1
The equation ∣z−1∣2+∣z+1∣2=4|z-1|^2+|z+1|^2=4 represents on the Argand plane
View answer and explanation
Correct option: D
A circle with centre origin and radius unity
Option D (A circle with centre origin and radius unity) is correct because the equation simplifies to ∣z∣=1|z|=1, which represents a circle centered at the origin with radius 1.

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16 Easy
16 of 979
Understanding +4 | -1
The locus represented by the equation ∣z−1∣=∣z−i∣\left| {z - 1} \right| = \left| {z - i} \right| is
View answer and explanation
Correct option: C
A line through the origin
Option C (A line through the origin) is correct because the equation represents the perpendicular bisector of the line segment joining 1 and ii, which passes through the origin.
17 Easy
17 of 979
Analyzing +4 | -1
The greatest number among 9,  3114,  176\sqrt[3]{{9,\;}}\sqrt[4]{{11}},\;\sqrt[6]{{17}} is
View answer and explanation
Correct option: A
93\sqrt[3]{9}
Option A is correct because when expressed with a common exponent, 93\sqrt[3]{9} has the largest base (6561) compared to the other options.
18 Medium
18 of 979
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.
19 Medium
19 of 979
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.
20 Easy
20 of 979
Analyzing +4 | -1
If α=cos⁡(8π11)+isin⁡(8π11)\alpha=\cos \left(\frac{8 \pi}{11}\right)+i \sin \left(\frac{8 \pi}{11}\right), where i=−1i=\sqrt{-1}, then Re⁡(α+α2+α3+α4+α5)\operatorname{Re}\left(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5\right) is
View answer and explanation
Correct option: B
−12-\frac{1}{2}
Option B (-1/2) is correct because the real part of the sum of the given complex numbers simplifies to -1/2 using properties of roots of unity.
21 Easy
21 of 979
Understanding +4 | -1
1+i2+i4+i6+…+i2n1+i^2+i^4+i^6+\ldots+i^{2 n} is
View answer and explanation
Correct option: D
Can't be determined
Option D (Can't be determined) is correct because the sum depends on the value of n, specifically whether n is even or odd.
22 Medium
22 of 979
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.
23 Hard
23 of 979
Application +4 | -1
If 1,  a1,  a2,…,an−11,\;{a_1},\;{a_2}, \ldots ,{a_{n - 1}} are the nn roots of unity, then the value of (1−a1)(1−a2)(1−a3)…(1−an−1)\left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right)\left( {1 - {a_3}} \right) \ldots \left( {1 - {a_{n - 1}}} \right) is equal to
View answer and explanation
Correct option: C
nn
Option C (n) is correct because the expression is equal to the derivative of x^n - 1 evaluated at x=1, which equals n.
24 Easy
24 of 979
Analyzing +4 | -1
If z=x+iyz=x+i y satisfies ∣z∣−2=0|z|-2=0 and ∣z−i∣−∣z+5i∣|z-i|-|z+5 i| =0=0, then
View answer and explanation
Correct option: C
x+2y+4=0x+2 y+4=0
Option C (x+2y+4=0x+2 y+4=0) is correct because the equation represents the perpendicular bisector of the line segment joining (0,1) and (0,-5), and the intersection with |z|=2 gives the solution.
25 Medium
25 of 979
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.
26 Easy
26 of 979
Analyzing +4 | -1
Let A={θ∈(0,2π):1+2isin⁡θ1−isin⁡θ is purely imaginary}.A=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta} \text { is purely imaginary}\right\}.
Then the sum of the elements in AA is
View answer and explanation
Correct option: A
4π4 \pi
Option A, 4π4 \pi, is the correct sum of the elements in set A, which contains values of θ\theta that make the given expression purely imaginary. The values are π4,3π4,5π4,7π4\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}.
27 Medium
27 of 979
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.
28 Easy
28 of 979
Application +4 | -1
If nn is any positive integer, then the value of i4n+1−i4n−12\frac{i^{4 n+1}-i^{4 n-1}}{2} equals
View answer and explanation
Correct option: C
ii
Option C (ii) is correct because the expression simplifies to ii.
29 Easy
29 of 979
Application +4 | -1
The value of [1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9]3{\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} is
View answer and explanation
Correct option: A
−12(3−i)- \frac{1}{2}\left( {\sqrt 3 - i} \right)
Option A (−12(3−i)- \frac{1}{2}\left( {\sqrt 3 - i} \right)) is correct because it is the simplified form of the given complex number expression after applying De Moivre's theorem.
30 Medium
30 of 979
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.
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