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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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871 Easy
871 of 979
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.
872 Medium
872 of 979
Analyzing +4 | -1
Two conjugate complex numbers z1z_1 and z1‾\overline{z_1} represent two adjacent vertices of a polygon of n\mathrm{n} sides, whose centre is origin. If Im⁡(z1)Re⁡(zˉ1)=2−1\frac{\operatorname{Im}\left(z_1\right)}{\operatorname{Re}\left(\bar{z}_1\right)}=\sqrt{2}-1 then ratio of the areas of the circles inscribed and circumscribed the polygon is
View answer and explanation
Correct option: C
2+122\frac{\sqrt{2}+1}{2 \sqrt{2}}
Option C (2+122\frac{\sqrt{2}+1}{2 \sqrt{2}}) is correct because it represents the square of the cosine of π8\frac{\pi}{8}, which is the ratio of the inscribed to circumscribed circle areas.
873 Medium
873 of 979
Application +4 | -1
If zz is a complex number in the Argand plane, then the equation ∣z−2∣2+∣z+2∣2=8{\left| {z - 2} \right|^2} + {\left| {z + 2} \right|^2} = 8 represents
View answer and explanation
Correct option: D
A circle
The correct answer is D (A circle) because the simplified equation ∣z−2∣2+∣z+2∣2=8{\left| {z - 2} \right|^2} + {\left| {z + 2} \right|^2} = 8 results in the equation of a circle.
874 Easy
874 of 979
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.
875 Easy
875 of 979
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.
876 Easy
876 of 979
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.
877 Easy
877 of 979
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.
878 Easy
878 of 979
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}.
879 Medium
879 of 979
Application +4 | -1
If z=32+i2(i=−1)z=\frac{\sqrt{3}}{2}+\frac{i}{2}(i=\sqrt{-1}), then (1+iz+z5+iz8)9\left(1+i z+z^{5}+i z^{8}\right)^{9} is equal to:
View answer and explanation
Correct option: D
-1
Option D (-1) is correct because substituting z=32+i2z=\frac{\sqrt{3}}{2}+\frac{i}{2} into the expression and simplifying using properties of complex numbers and cube roots of unity leads to -1.
880 Easy
880 of 979
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.
881 Easy
881 of 979
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.
882 Medium
882 of 979
Application +4 | -1
If 3+i=(a+ib)(c+id)\sqrt{3}+i=(a+i b)(c+i d), then tan⁡−1ba+tan⁡−1dc\tan ^{-1} \frac{b}{a}+\tan ^{-1} \frac{d}{c} is equal to
View answer and explanation
Correct option: C
π6+nπ\frac{\pi}{6}+n \pi, for some integer nn
Option C (π6+nπ\frac{\pi}{6}+n \pi, for some integer nn) is correct because the argument of a product of complex numbers is the sum of their arguments.
883 Easy
883 of 979
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.
884 Hard
884 of 979
Application +4 | -1
If mm and xx are two real numbers, then e2micot⁡−1x(xi+1xi−1)me^{2 m i \cot ^{-1} x}\left(\frac{x i+1}{x i-1}\right)^m (where i=−1i=\sqrt{-1} ) is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because after simplification using trigonometric identities and Euler's formula, the expression equals 1.
885 Easy
885 of 979
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.
886 Easy
886 of 979
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.
887 Easy
887 of 979
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.
888 Medium
888 of 979
Application +4 | -1
For complex numbers z1=x1+iyz_1=x_1+i y, and z2z_2 =x2+iy2=x_2+i y_2, we write z1∩z2z_1 \cap z_2 if x1≤x2x_1 \leq x_2 and y1≤y2y_1 \leq y_2. Let zz be a complex number such that 1∩z1 \cap z, then
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Correct option: C
1−z1+z∩0\frac{1-z}{1+z} \cap 0
Option C (1−z1+z∩0\frac{{1-z}}{{1+z}} \cap 0) is correct because, given 1∩z1 \cap z, the real and imaginary parts of 1−z1+z\frac{{1-z}}{{1+z}} are both non-positive.
889 Medium
889 of 979
Application +4 | -1
If the 5th 5^{\text {th }} roots of unity are 1,z1,z2,z3,z41, z_1, z_2, z_3, z_4 then z12+z22+z32+z42z_1{ }^2+z_2{ }^2+z_3{ }^2+z_4{ }^2 is equal to
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Correct option: B
-1
Option B (-1) is correct because the sum of the squares of the non-real fifth roots of unity is -1.
890 Easy
890 of 979
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}|.
891 Medium
891 of 979
Application +4 | -1
Let z1z_1 and z2z_2 be nnth roots of unity which subtend a right angle at the origin. Then nn must be of the form
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Correct option: D
4k4 k
Option D (4k4k) is correct because the angle between the nth roots of unity must be a right angle, implying that n is a multiple of 4.
892 Medium
892 of 979
Application +4 | -1
For any two complex numbers z1{z_1} and z2{z_2} and any real numbers aa and bb, ∣az1−bz2∣2+∣bz1+az2∣2{\left| {a{z_1} - b{z_2}} \right|^2} + {\left| {b{z_1} + a{z_2}} \right|^2} is equal to
View answer and explanation
Correct option: B
(a2+b2)(∣z1∣2+∣z2∣2)\left( {{a^2} + {b^2}} \right)\left( {{{\left| {{z_1}} \right|}^2} + {{\left| {{z_2}} \right|}^2}} \right)
Option B ((a2+b2)(∣z1∣2+∣z2∣2)\left( {{a^2} + {b^2}} \right)\left( {{{\left| {{z_1}} \right|}^2} + {{\left| {{z_2}} \right|}^2}} \right)) is correct because expanding and simplifying the expression leads to this result.
893 Medium
893 of 979
Application +4 | -1
Let ωn=cos⁡(2πn)+isin⁡(2πn),  i2=−1,{\omega _n} = \cos \left( {\frac{{2\pi }}{n}} \right) + i\sin \left( {\frac{{2\pi }}{n}} \right),\;{i^2} = - 1, then (x+yω3+zω32)(x+yω32+zω3)\left( {x + y{\omega _3} + z\omega _3^2} \right)\left( {x + y\omega _3^2 + z{\omega _3}} \right) is equal to
View answer and explanation
Correct option: C
x2+y2+z2−yz−zx−xy{x^2} + {y^2} + {z^2} - yz - zx - xy
Option C (x2+y2+z2−yz−zx−xy{x^2} + {y^2} + {z^2} - yz - zx - xy) is correct because substituting ω3=ω\omega_3 = \omega and ω32=ω2\omega_3^2 = \omega^2 (where ω\omega is a cube root of unity) and expanding the product leads to the given expression.
894 Easy
894 of 979
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.
895 Easy
895 of 979
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.
896 Medium
896 of 979
Application +4 | -1
Locus of point zz so that z,iz, i, and izi z are collinear, is
View answer and explanation
Correct option: B
a circle
Option B (a circle) is correct because the condition for collinearity leads to an equation that represents a circle in the complex plane.
897 Medium
897 of 979
Application +4 | -1
Let α,  β\alpha ,\;\beta be the roots of x2−2xcos⁡ϕ+1=0{x^2} - 2x\cos \phi + 1 = 0, then the equation whose roots are αn,  βn,  {\alpha ^n},\;{\beta ^n},\; is
View answer and explanation
Correct option: B
x2−2xcos⁡nϕ+1=0{x^2} - 2x\cos n\phi + 1 = 0
Option B (x2−2xcos⁡nϕ+1=0{x^2} - 2x\cos n\phi + 1 = 0) is correct because the roots can be expressed using Euler's formula, and raising them to the power of n results in a new equation with coefficients involving cos(nφ).
898 Easy
898 of 979
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.
899 Easy
899 of 979
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.
900 Easy
900 of 979
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.
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