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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Easy

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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 61-90 of 612 questions

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61 Easy
61 of 612
Understanding +4 | -1
The additive inverse of 1−i1 - i is
View answer and explanation
Correct option: B
−1+i- 1 + i
Option B (-1 + i) is correct because it is the additive inverse that, when added to the original number, results in zero.
62 Easy
62 of 612
Application +4 | -1
If ω(≠1)\omega \left( { \ne 1} \right) be a cube root of unity and (1+ω)7=A+Bω,{\left( {1 + \omega } \right)^7} = A + B\omega , then AA and BB are respectively the numbers:
View answer and explanation
Correct option: B
1,  11,\;1
Option B (1, 1) is correct because simplifying the expression using properties of cube roots of unity leads to A = 1 and B = 1.
63 Easy
63 of 612
Application +4 | -1
If ∣6i−3i143i−1203i∣=x+iy,\left| {\begin{array}{cccccccccccccccccccc} {6i}&{ - 3i}&1\\ 4&{3i}&{ - 1}\\ {20}&3&i \end{array}} \right| = x + iy, then
View answer and explanation
Correct option: D
x=0,y=0x = 0,y = 0
Option D (x=0,y=0x = 0, y = 0) is correct because the determinant is zero after applying column operations.
64 Easy
64 of 612
Application +4 | -1
If zz and ww are two complex numbers such that ∣zw∣=1\left| {zw} \right| = 1 and arg (z) - arg (w)=π2(w) = \frac{\pi }{2}, then
View answer and explanation
Correct option: A
zˉw=−i\bar zw = - i
Option A (zˉw=−i\bar zw = - i) is correct because, given the conditions on the magnitudes and arguments of z and w, the expression simplifies to -i.
65 Easy
65 of 612
Application +4 | -1
The area of the triangle formed by the points representing −z,  iz- z,\;iz and z−izz - iz in the Argand plane is
View answer and explanation
Correct option: C
32∣z∣2\frac{3}{2}{\left| z \right|^2}
The correct answer is C (32∣z∣2\frac{3}{2}{\left| z \right|^2}) because the area of the triangle formed by the given points in the Argand plane is 32\frac{3}{2} times the square of the modulus of z.
66 Easy
66 of 612
Application +4 | -1
The region represented by {z=x+iy∈C:∣z∣−Re(z)≤1}\{ z = x + iy \in C:\left| z \right| - {\mathop\mathrm{Re}\nolimits} (z) \le 1\} is also given by the inequality
View answer and explanation
Correct option: B
y2≤2(x+12){y^2} \le 2(x + \frac{1}{2})
Option B (y2≤2(x+12){y^2} \le 2(x + \frac{1}{2})) is correct because it represents the inequality derived from the given condition ∣z∣−Re(z)≤1|z| - Re(z) \le 1 after substituting z=x+iyz = x + iy and simplifying.
67 Easy
67 of 612
Application +4 | -1
The equation ∣z−i∣=∣z−1∣,i=−1,\left| {z - i} \right| = \left| {z - 1} \right|,i = \sqrt { - 1} , represents
View answer and explanation
Correct option: B
The line passing through the origin with slope 1
Option B (The line passing through the origin with slope 1) is correct because the equation simplifies to y = x, which represents a line through the origin with a slope of 1.
68 Easy
68 of 612
Analyzing +4 | -1
If the roots of the equation 1x+p+1x+q=1r,  (x≠−p,  x≠−q,  r≠0)\frac{1}{{x + p}} + \frac{1}{{x + q}} = \frac{1}{r},\;\left( {x \ne - p,\;x \ne - q,\;r \ne 0} \right) are equal in magnitude but opposite in sign, then p+qp+q is equal to
View answer and explanation
Correct option: B
2r2r
Option B (2r) is correct because when the roots are equal in magnitude but opposite in sign, the coefficient of x in the quadratic equation derived from the given equation must be zero, leading to p + q = 2r.
69 Easy
69 of 612
Application +4 | -1
If z=3+iz = \sqrt 3 + i, then the argument of z2ez−i{z^2}{e^{z - i}} is equal to
View answer and explanation
Correct option: D
π/3\pi /3
Option D (π/3\pi /3) is correct because the argument of the complex number z2ez−i{z^2}{e^{z - i}} simplifies to π3\frac{\pi}{3} when z=3+iz = \sqrt{3} + i.
70 Easy
70 of 612
Application +4 | -1
If x=3+ix = 3 + i, then x3−3x2−8x+15{x^3} - 3{x^2} - 8x + 15 is equal to
View answer and explanation
Correct option: B
−15- 15
Option A (3) is correct because for a quadratic equation to have equal roots, the discriminant must be zero, and solving for 'a' gives a = 3.
71 Easy
71 of 612
Analyzing +4 | -1
If ∣z−2∣=min⁡{∣z−1∣,  ∣z−5∣},    \left| {z - 2} \right| = \min \left\{ {\left| {z - 1} \right|,\;\left| {z - 5} \right|} \right\},\;\; where zz is a complex number, then
View answer and explanation
Correct option: C
Re(z)∈{32,72}{\mathop\mathrm{Re}\nolimits} \left( z \right) \in \left\{ {\frac{3}{2},\frac{7}{2}} \right\}
Option C is correct because the real part of z can be either 3/2 or 7/2, satisfying the given condition based on the minimum distance.
72 Easy
72 of 612
Application +4 | -1
If ∏p=1reipθ=1,\prod\limits_{p = 1}^r {{e^{ip\theta }} = 1,} where ∏\prod denotes the continued product and i=−1i=\sqrt{-1}, the most general value of θ\theta is
(where, nn is an integer)
View answer and explanation
Correct option: D
4nπr(r+1),n∈I\frac{4 n \pi}{r(r+1)}, n \in I
Option D (4nπr(r+1),n∈I\frac{{4 n \pi }}{{r(r + 1)}}, n \in I) is correct because it correctly sums the geometric series in the exponent and solves for θ\theta using the properties of complex exponentials.
73 Easy
73 of 612
Analyzing +4 | -1
If zεCz \varepsilon \mathbf{C}, the minimum value of ∣z∣+∣z−i∣|z|+|z-i| is attained at
View answer and explanation
Correct option: C
infinite number of points
Option C (infinite number of points) is correct because the minimum value is attained for all points on the line segment joining 0 and i, which represents an infinite number of points.
74 Easy
74 of 612
Application +4 | -1
If α,β\alpha, \beta are the complex cube roots of unity, then α3+β3+α−2β−2=\alpha^3+\beta^3+\alpha^{-2} \beta^{-2}=
View answer and explanation
Correct option: B
3
The correct answer is 3 because when α\alpha and β\beta are complex cube roots of unity, substituting ω\omega and ω2\omega^2 simplifies the expression to 3.
75 Easy
75 of 612
Analyzing +4 | -1
The value of sum ∑n=113(in+in+1),    where   i=−1,  \sum \limits_{n = 1}^{13} \left( {{i^n} + {i^{n + 1}}} \right),\;\;\mathrm{where}\,\;i = \sqrt { - 1} ,\; equals
View answer and explanation
Correct option: B
i−1i - 1
Option B (i−1i - 1) is correct because the sum telescopes, leaving only the first two terms.
76 Easy
76 of 612
Analyzing +4 | -1
If ω\omega is complex cube root of unity and a,b,ca, b, c are such that 1a+ω+1b+ω+1c+ω=2ω2\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2 \omega^2 and 1a+ω2+1b+ω2+1c+ω2=2ω\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2 \omega, then the value of 1a+1+1b+1+1c+1\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} is equal to
View answer and explanation
Correct option: D
22
Option D (2) is correct because by recognizing omega and omega squared as roots of an equation, the third root can be found and used to evaluate the expression.
77 Easy
77 of 612
Understanding +4 | -1
If ∣z+8∣+∣z−8∣=16\left| {z + 8} \right| + \left| {z - 8} \right| = 16, where zz is a complex number, then the point zz will lie on
View answer and explanation
Correct option: C
A straight line
Option C (A straight line) is correct because the equation represents the locus of a point where the sum of its distances from two fixed points is constant, which degenerates into a straight line when the constant equals the distance between the fixed points.
78 Easy
78 of 612
Application +4 | -1
If z=(13+12i)7+(13−12i)7z=\left(\frac{1}{\sqrt{3}}+\frac{1}{2} i\right)^7+\left(\frac{1}{\sqrt{3}}-\frac{1}{2} i\right)^7, then
View answer and explanation
Correct option: B
Im⁡(z)=0\operatorname{Im}(z)=0
Option B (Im(z)=0) is correct because the expression is the sum of a complex number raised to the 7th power and its conjugate raised to the 7th power, resulting in a real number, hence the imaginary part is zero.
79 Easy
79 of 612
Analyzing +4 | -1
If z+1z=1z+\frac{1}{z}=1 and a=z2017+1z2017a=z^{2017}+\frac{1}{z^{2017}} and bb is the last digit of the number 22n−12^{2^n}-1, when the integer n>1n>1, the value of a2+b2a^2+b^2 is
View answer and explanation
Correct option: C
26
Option C (26) is correct because by solving the quadratic equation and using properties of cube root of unity, and last digit properties, the value of the expression can be found.
80 Easy
80 of 612
Application +4 | -1
If z1=10+6i,z2=4+6iz_1=10+6 i, z_2=4+6 i and zz is a complex number such that amp⁡(z−z1z−z2)=π4\operatorname{amp}\left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4}, then the value of ∣z−7−9i∣|z-7-9 i| is equal to
View answer and explanation
Correct option: A
323 \sqrt{2} \text{}
Option A (323 \sqrt{2} \text{}) is correct because it represents the magnitude of the complex number |z - 7 - 9i| after simplifying the given amplitude condition.
81 Easy
81 of 612
Application +4 | -1
If 2x.3x+4=7x{2^x}{.3^{x + 4}} = {7^x}, then xx is equal to
View answer and explanation
Correct option: A
4log⁡e3log⁡e7−log⁡e6\frac{{4{{\log }_e}3}}{{{{\log }_e}7 - {{\log }_e}6}}
Option A (4log⁡e3log⁡e7−log⁡e6\frac{{4{{\log }_e}3}}{{{{\log }_e}7 - {{\log }_e}6}}) is correct because it is obtained by taking the logarithm of both sides of the equation and solving for x.
82 Easy
82 of 612
Application +4 | -1
An equation of straight line joining the complex numbers aa and ibi b (where a,bRa, b \boldsymbol{R} and a,b≠0a, b \neq 0 ) is
View answer and explanation
Correct option: A
z(1a−ib)+zˉ(1a+ib)=2z\left(\frac{1}{a}-\frac{i}{b}\right)+\bar{z}\left(\frac{1}{a}+\frac{i}{b}\right)=2
Option A (z(1a−ib)+zˉ(1a+ib)=2z(\frac{{1}}{{a}}-\frac{{i}}{{b}})+\bar{{z}}(\frac{{1}}{{a}}+\frac{{i}}{{b}})=2) is correct because it represents the equation of a line passing through the complex numbers aa and ibib.
83 Easy
83 of 612
Application +4 | -1
If ∣z∣=3|z|=3, the area of the triangle whose sides are z,ωεz, \omega \varepsilon and z+ωεz+\omega \varepsilon (where ω\omega is a complex cube root of unity) is
View answer and explanation
Correct option: A
93/49 \sqrt{3} / 4
Option A (93/49 \sqrt{{3}} / 4) is correct because the triangle formed by zz, ωz\omega z, and z+ωzz + \omega z is equilateral with side length ∣z∣=3|z| = 3, and the area of an equilateral triangle is 34×side2\frac{{\sqrt{{3}}}}{{4}} \times side^2.
84 Easy
84 of 612
Analyzing +4 | -1
The value of the expression
(2−ω)(2−ω2)+2(3−ω)(3−ω2)+.……+.…+(n−1)(n−ω)(n−ω2)\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) + . \ldots \ldots + . \ldots + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)
Where ω\omega is an imaginary cube root of unit is
View answer and explanation
Correct option: A
(n−1)n(n2+3n+4)/4\left( {n - 1} \right)n\left( {{n^2} + 3n + 4} \right)/4
Option A is correct because the expression simplifies to (n−1)n(n2+3n+4)4\frac{{(n - 1)n(n^2 + 3n + 4)}}{{4}} after expanding and summing the terms using the properties of cube roots of unity.
85 Easy
85 of 612
Application +4 | -1
The equation of the locus of zz such that ∣z−iz+i∣=2\left| {\frac{{z - i}}{{z + i}}} \right| = 2, where z=x+iyz = x + iy is a complex number, is
View answer and explanation
Correct option: B
3x2+3y2+10y+3=03{x^2} + 3{y^2} + 10y + 3 = 0
The correct answer is B (3x2+3y2+10y+3=03{x^2} + 3{y^2} + 10y + 3 = 0) because substituting z=x+iyz = x + iy and simplifying the given equation leads to this equation of a circle.
86 Easy
86 of 612
Application +4 | -1
If a,b,x,y∈R,ω≠1a, b, x, y \in \mathbf{R}, \omega \neq 1, is a cube root of unity and (a+bω)7=x+yω(a+b \omega)^7=x+y \omega, then (b+aω)7(b+a \omega)^7 equals
View answer and explanation
Correct option: A
y+xωy+x \omega
Option A (y+xωy+x \omega) is correct because by taking the conjugate and manipulating the expression using the properties of cube roots of unity, we find that (b+aω)7=y+xω(b + a\omega)^7 = y + x\omega.
87 Easy
87 of 612
Application +4 | -1
If x+iy=a+ibc+idx+i y=\sqrt{\frac{a+i b}{c+i d}}, then the value of (x2+y2)2\left(x^2+y^2\right)^2 is
View answer and explanation
Correct option: A
a2+b2c2+d2\frac{a^2+b^2}{c^2+d^2}
Option A (a2+b2c2+d2\frac{a^2+b^2}{c^2+d^2}) is correct because multiplying the complex number by its conjugate and squaring the result gives the correct expression.
88 Easy
88 of 612
Application +4 | -1
The value of sum ∑n=113(in+in+1)\sum \limits_{n = 1}^{13} \left( {{i^n} + {i^{n + 1}}} \right), where i=−1i = \sqrt { - 1}, equals
View answer and explanation
Correct option: B
i−1i - 1
Option B (i−1i - 1) is correct because the sum simplifies to i13+i14i^{13} + i^{14} which equals i−1i - 1 due to the cyclic nature of powers of i.
89 Easy
89 of 612
Application +4 | -1
The complex number zz, satisfies the condition ∣z−25z∣=24\left|z-\frac{25}{z}\right|=24. The maximum distance from the origin of coordinates to the point zz, is
View answer and explanation
Correct option: A
25
Option A (25) is correct because the maximum distance from the origin to the point z is found by solving the inequality derived from the given condition using properties of complex numbers.
90 Easy
90 of 612
Application +4 | -1
If z∈Cz \in \mathbf{C}, then Re⁡(zˉ2)=k2,k>0\operatorname{Re}\left(\bar{z}^2\right)=k^2, k>0, represents
View answer and explanation
Correct option: D
a hyperbola
Option D (a hyperbola) is correct because the equation Re⁡(zˉ2)=k2\operatorname{Re}(\bar{z}^2) = k^2 simplifies to x2−y2=k2x^2 - y^2 = k^2, which is the equation of a hyperbola.
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