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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Easy

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Practice format and chapter quiz

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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541 Easy
541 of 612
Application +4 | -1
The value of (cos⁡π2+isin⁡π2)(cos⁡π4+isin⁡π4)(cos⁡π8+  isin⁡π8)…∞\left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)\left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)\left( {\cos \frac{\pi }{8} + \;i\sin \frac{\pi }{8}} \right) \ldots \infty is
View answer and explanation
Correct option: C
−1-1
Option C (-1) is correct. The infinite product of complex numbers in polar form simplifies to cos(π) + isin(π), which equals -1.
542 Easy
542 of 612
Application +4 | -1
If m1,m2,m3{m_1},{m_2},{m_3} and m4{m_4} respectively denote the moduli of the complex numbers 1+4i,  3+i,  1−i1 + 4i,\;3 + i,\;1 - i and 2−3i2 - 3i then the correct one, among the following is
View answer and explanation
Correct option: C
m3<m2<m4<m1{m_3} < {m_2} < {m_4} < {m_1}
The correct answer is C (m3<m2<m4<m1{m_3} < {m_2} < {m_4} < {m_1}) because calculating the moduli and comparing them yields this order.
543 Easy
543 of 612
Application +4 | -1
If zz is a complex number, then z2+zˉ2=2z^2+\bar{z}^2=2 represents
View answer and explanation
Correct option: C
a hyperbola
Option C (a hyperbola) is correct because the equation z2+zˉ2=2z^2 + \bar{z}^2 = 2 simplifies to x2−y2=1x^2 - y^2 = 1, which represents a hyperbola.
544 Easy
544 of 612
Application +4 | -1
The argument of 1+i31−i3\frac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }} is
View answer and explanation
Correct option: A
2π/32\pi /3
Option A (2π/32\pi /3) is correct because after simplifying the complex number, its argument falls in the second quadrant with a reference angle of π/3\pi/3.
545 Easy
545 of 612
Analyzing +4 | -1
If the number (1−i)n(1+i)n−2\frac{(1-i)^n}{(1+i)^{n-2}} is real and positive, then nn is
View answer and explanation
Correct option: C
4λ+14 \lambda+1
Option C (4λ+14 \lambda+1) is correct because the expression (1−i)n(1+i)n−2\frac{{(1-i)^n}}{{(1+i)^{n-2}}} is real and positive only when nn is of the form 4λ+14 \lambda+1, where λ\lambda is an integer.
546 Easy
546 of 612
Application +4 | -1
If z∈Cz \in \mathbf{C} and 2z=∣z∣+i2 z=|z|+i, then zz equals
View answer and explanation
Correct option: A
36+12i\frac{\sqrt{3}}{6}+\frac{1}{2} i
Option C (36+14i\frac{{\sqrt{{3}}}}{{6}} + \frac{{1}}{{4}}i) is incorrect. The correct answer is not among the options provided. The correct solution involves equating real and imaginary parts and solving for z, which results in z = 1/(23\sqrt{3}) + i/2.
547 Easy
547 of 612
Application +4 | -1
If (1−i)n=2n(1-i)^n=2^n, then the value of nn is
View answer and explanation
Correct option: A
0
Option A (0) is correct because taking the modulus of both sides of the equation and solving for n results in n = 0.
548 Easy
548 of 612
Application +4 | -1
If ω\omega is a complex cube root of unity, then the value of ω99+ω100+ω101{\omega ^{99}} + {\omega ^{100}} + {\omega ^{101}} is
View answer and explanation
Correct option: D
0
Option D (0) is correct because the sum of consecutive powers of omega (ω\omega) where the powers differ by 1, and are multiples of 3, is always 0.
549 Easy
549 of 612
Application +4 | -1
If nn is a positive integer, then (1+i3)n+(1−i3)n{\left( {1 + i\sqrt 3 } \right)^n} + {\left( {1 - i\sqrt 3 } \right)^n} is equal to
View answer and explanation
Correct option: C
2n+1cos⁡nπ3{2^{n + 1}}\cos \frac{{n\pi }}{3}
Option C (2n+1cos⁡nπ3{2^{n + 1}}\cos \frac{{n\pi }}{3}) is correct because by converting the complex numbers to polar form and applying De Moivre's theorem, the expression simplifies to 2n+1cos⁡nπ3{2^{n + 1}}\cos \frac{{n\pi }}{3}.
550 Easy
550 of 612
Application +4 | -1
If z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i\sin \theta ), then the value of zzˉ+zˉz\frac{z}{{\bar z}} + \frac{{\bar z}}{z} is
View answer and explanation
Correct option: B
2cos⁡2θ2\cos 2\theta
Option B (2cos⁡2θ2\cos 2\theta) is correct because it is the simplified form of the given expression after substituting z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i\sin \theta ) and using Euler's formula.
551 Easy
551 of 612
Application +4 | -1
If ∣z+4∣=3|z+4|=3, then greatest possible value of ∣z−7∣/7|z-7| / 7 is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the maximum value of ∣z−7∣|z-7| occurs when z is farthest from 7 on the circle, which leads to a value of 2.
552 Easy
552 of 612
Application +4 | -1
A particle PP starts from the point z0=1+2i,i=−1z_0=1+2 i, i=\sqrt{-1}. It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point z1z_1. From z1z_1, the particle moves 2\sqrt{2} units in the direction of the vector i^+j^\hat{\mathbf{i}}+\hat{\mathbf{j}} and then it moves through an angle π2\frac{\pi}{2} in anti-clockwise direction on a circle with centre at origin, to reach a point z2z_2, then the point z2z_2 is given by
View answer and explanation
Correct option: D
−6+7i-6+7 i
Option D (-6+7 i) is correct because it represents the final position of the particle after all the described movements.
553 Easy
553 of 612
Understanding +4 | -1
If ∣z1+z2∣=∣z1∣+∣z2∣\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|, then one of the values of arg⁡(z2/z1)\arg \left(z_2 / z_1\right) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because if ∣z1+z2∣=∣z1∣+∣z2∣|z_1 + z_2| = |z_1| + |z_2|, then z1z_1 and z2z_2 have the same argument, so arg⁡(z2/z1)=0\arg(z_2/z_1) = 0.
554 Easy
554 of 612
Application +4 | -1
If zz is a non-real 7th{7^{th}} root of −1- 1, then z86+z175+z289{z^{86}} + {z^{175}} + {z^{289}} is equal to
View answer and explanation
Correct option: B
−1- 1
Option B (-1) is correct because after simplifying the powers of z using the fact that z7 = -1, the expression reduces to -1.
555 Easy
555 of 612
Application +4 | -1
If ∣z+iz−i∣=3\left| {\frac{{z + i}}{{z - i}}} \right| = \sqrt 3, then radius of the circle is
View answer and explanation
Correct option: C
3\sqrt 3
Option C (3\sqrt 3) is correct because the equation of the circle can be derived from the given modulus condition, and the radius can be found from the equation of the circle.
556 Easy
556 of 612
Application +4 | -1
If α,β\alpha, \beta are distinct complex numbers with ∣β∣=1|\beta|=1, then value of ∣β−α1−αˉβ∣\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right| equals
View answer and explanation
Correct option: A
1
Option A (1) is correct because the expression simplifies to 1 using the property that ∣β∣=1|\beta| = 1 and ∣z∣=∣zˉ∣|z| = |\bar{z}|.
557 Easy
557 of 612
Application +4 | -1
If ω(≠1)\omega(\neq 1) is a cube root of unity and (1+ω)7=A+Bω(1+\omega)^7=A+B \omega, then (A,B)(A, B) equals to
View answer and explanation
Correct option: A
(1,1)(1,1)
Option A ((1,1)) is correct because simplifying the expression (1+ω)7 using the properties of cube roots of unity leads to 1+ω, thus A=1 and B=1.
558 Easy
558 of 612
Application +4 | -1
If the cube roots of unity are 1,ω,ω21, \omega, \omega^2, then the roots of the equation x3−3x2+3x+63=0x^3-3 x^2+3 x+63=0 are
View answer and explanation
Correct option: C
−3,1−4ω,1−4ω2-3,1-4 \omega, 1-4 \omega^2
Option D (−3,1+4ω,1+4ω2-3,1+4 \omega, 1+4 \omega^2) is correct because by substituting x=y+1x = y + 1 into the equation, we can solve for y and then find the roots in terms of ω\omega.
559 Easy
559 of 612
Application +4 | -1
If 3(x−2)5≥5(2−x)3,\frac{{3\left( {x - 2} \right)}}{5} \ge \frac{{5\left( {2 - x} \right)}}{3}, then xx belongs to the interval
View answer and explanation
Correct option: B
[2,∞)\left[ {2,\infty } \right)
Option B ([2,∞)) is the correct interval because solving the inequality leads to x being greater than or equal to 2.
560 Easy
560 of 612
Application +4 | -1
If i=−1i = \sqrt { - 1}, then 4+5(−12+i32)334+3(−12+i32)3654 + 5{\left( { - \frac{1}{2} + \frac{{i\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \frac{1}{2} + \frac{{i\sqrt 3 }}{2}} \right)^{365}} is equal to
View answer and explanation
Correct option: C
i3i\sqrt 3
The correct answer is C (i3i\sqrt 3) because, after substituting ω\omega and simplifying using the properties of complex cube roots of unity, the expression equals i3i\sqrt 3.
561 Easy
561 of 612
Application +4 | -1
The area of the triangle whose vertices are i,ωi,\omega and ω2,{\omega ^2},, where i=−1i = \sqrt { - 1} and ω,ω2\omega ,{\omega ^2} are complex cube roots of unity, is
View answer and explanation
Correct option: D
34\frac{{\sqrt 3 }}{4}
Option D (34\frac{{\sqrt 3 }}{4}) is correct because it is the area of the triangle calculated using the determinant formula with the given vertices.
562 Easy
562 of 612
Application +4 | -1
If ∣z∣=1|z|=1 and z≠±1z \neq \pm 1, then all the values of z1−z2\frac{z}{1-z^2} lie on
View answer and explanation
Correct option: D
the YY-axis
Option D (the YY-axis) is correct because the expression simplifies to a purely imaginary number, indicating that all values lie on the Y-axis.
563 Easy
563 of 612
Analyzing +4 | -1
If (1+i)(1+2i)(1+3i)…(1+ni)=α+iβ(1+i)(1+2 i)(1+3 i) \ldots(1+n i)=\alpha+i \beta, then 2⋅5⋅10…(1+n2)2 \cdot 5 \cdot 10 \ldots\left(1+n^2\right) is equal to
View answer and explanation
Correct option: C
α2+β2\alpha^2+\beta^2
Option C (α2+β2\alpha^2+\beta^2) is correct because taking the modulus squared of both sides of the given equation leads to 2⋅5⋅10…(1+n2)=α2+β22 \cdot 5 \cdot 10 \ldots (1+n^2) = \alpha^2 + \beta^2.
564 Easy
564 of 612
Analyzing +4 | -1
If z2+z+1=0z^2+z+1=0, where zz is a complex number, then values of S=(z+1z)2+(z2+1z2)2+(z3+1z3)2+⋯+(z6+1z6)2 is S=\left(z+\frac{1}{z}\right)^2+\left(z^2+\frac{1}{z^2}\right)^2+\left(z^3+\frac{1}{z^3}\right)^2+\cdots+\left(z^6+\frac{1}{z^6}\right)^2 \text { is }
View answer and explanation
Correct option: A
12
Option A (12) is correct because by substituting z=ωz = \omega (a complex cube root of unity) into the expression and using the properties of ω\omega, the sum simplifies to 12.
565 Easy
565 of 612
Application +4 | -1
The value of log⁡2log⁡3…log⁡1001009998..21{\log _2}{\log _3} \ldots {\log _{100}}{100^{{{99}^{{{98}^{{.^{{.^{{2^1}}}}}}}}}}} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because the expression simplifies down to log base 2 of 2, which equals 1.
566 Easy
566 of 612
Application +4 | -1
Which of the following is a fourth root of 12+i32\frac{1}{2} + i\frac{{\sqrt 3 }}{2} ?
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Correct option: A
cisπ12\mathrm{cis}\frac{\pi }{{12}}
Option A (cisπ12\mathrm{cis}\frac{\pi }{{12}}) is correct because it represents the fourth root of the given complex number in polar form.
567 Easy
567 of 612
Application +4 | -1
If z=12−2iz=\frac{1}{2}-2 i is such that ∣z+1∣=αz+β(1+i)|z+1|=\alpha z+\beta(1+i), i=−1i=\sqrt{-1} and α,β∈R\alpha, \beta \in {R}, then α+β\alpha+\beta is equal to
View answer and explanation
Correct option: D
3
Option D (3) is correct because solving the equation |z+1|=αz+β(1+i) for α and β yields α=1 and β=2, so α+β=3.
568 Easy
568 of 612
Understanding +4 | -1
The points representing complex number zz for which ∣z−3∣=∣z−5∣\left| {z - 3} \right| = \left| {z - 5} \right| lie on the locus given by
View answer and explanation
Correct option: C
A straight line
Option D (Perpendicular bisector of line joining z1 and z2) is correct because the given condition implies that the point P is equidistant from z1 and z2, which defines the perpendicular bisector.
569 Easy
569 of 612
Analyzing +4 | -1
If ∣z1∣=∣z2∣=∣z3∣=1\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1 and z1+z2+z3=0z_1+z_2+z_3=0, then area of the triangle whose vertices are z1,z2,z3z_1, z_2, z_3 is
View answer and explanation
Correct option: A
33/43 \sqrt{3} / 4
Option A (33/43 \sqrt{3} / 4) is correct because the given conditions imply that the vertices form an equilateral triangle with side 3\sqrt{3}, and thus the area is 33/43 \sqrt{3} / 4.
570 Easy
570 of 612
Analyzing +4 | -1
If z1,  z2,  z3,  z4{z_1},\;{z_2},\;{z_3},\;{z_4} are four complex numbers represented by the vertices of a quadrilateral taken in order such that z1−z4=z2−z3    and  arg⁡(z4−z1z2−z1)=π2,  {z_1} - {z_4} = {z_2} - {z_3}\;\;\mathrm{and}\;\arg \left( {\frac{{{z_4} - {z_1}}}{{{z_2} - {z_1}}}} \right) = \frac{\pi }{2},\; then the quadrilateral is
View answer and explanation
Correct option: B
A rectangle
Option B is correct because the quadrilateral is a rectangle, as the diagonals bisect each other and one angle is π2\frac{\pi}{2}.
Revision notes and key takeaways

Review COMPLEX NUMBERS in your textbook or class notes, then use this set to check your understanding.

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