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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 31-60 of 612 questions
31 Easy
31 of 612
Application +4 | -1
The locus of the points zz which satisfy the condition arg⁡(z−1z+1)=π3\arg \left( {\frac{{z - 1}}{{z + 1}}} \right) = \frac{\pi }{3} is
View answer and explanation
Correct option: B
A circle
Option B (A circle) is correct because the equation derived from the given condition arg⁡(z−1z+1)=π3\arg \left( {\frac{{z - 1}}{{z + 1}}} \right) = \frac{\pi }{3} represents a circle.
32 Easy
32 of 612
Application +4 | -1
If α\alpha and β\beta are imaginary cube roots of unity, then α4+β4+1αβ{\alpha ^4} + {\beta ^4} + \frac{1}{{\alpha \beta }} is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because using the properties of cube roots of unity, the expression simplifies to zero.
33 Easy
33 of 612
Application +4 | -1
If ω\omega is an imaginary cube root of unity, nn is a positive integer but not a multiple of 3, then the value of 1+ωn+ω2n1 + {\omega ^n} + {\omega ^{2n}} is
View answer and explanation
Correct option: C
0
Option C (0) is correct because when n is not a multiple of 3, the expression 1+ωn+ω2n1 + {\omega ^n} + {\omega ^{2n}} simplifies to 0, where ω\omega is a complex cube root of unity.
34 Easy
34 of 612
Analyzing +4 | -1
The area of the polygon, whose vertices are the non-real roots of the equation z‾=iz2\overline z = i{z^2} is
View answer and explanation
Correct option: A
334\frac{{3\sqrt 3 }}{4}
Option A (334\frac{{3\sqrt 3 }}{4}) is correct because solving the equation z‾=iz2\overline z = i{z^2} gives the vertices of a polygon, and calculating the area of the polygon results in 334\frac{{3\sqrt 3 }}{4}.
35 Easy
35 of 612
Application +4 | -1
The smallest positive integer nn for which (1+i1−i)n=1{\left( {\frac{{1 + i}}{{1 - i}}} \right)^n} = 1, is
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the smallest positive integer n for which the given expression equals 1 is 4, not any of the provided options.
36 Easy
36 of 612
Analyzing +4 | -1
If mm and nn are the smallest positive integers satisfying the relation (2CiS⁡π6)m=(4CiS⁡π4)n\left(2 \operatorname{CiS} \frac{\pi}{6}\right)^m=\left(4 \operatorname{CiS} \frac{\pi}{4}\right)^n, where i=−1i=\sqrt{-1}, (m+n)(m+n) equals to
View answer and explanation
Correct option: B
72
Option B (72) is correct because simplifying the given expression and equating the powers leads to m=48 and n=24, thus m+n=72.
37 Easy
37 of 612
Analyzing +4 | -1
If x=91/3⋅91/9⋅91/27…∞,y=41/3⋅4−1/9⋅41/27…∞x=9^{1 / 3} \cdot 9^{1 / 9} \cdot 9^{1 / 27} \ldots \infty, y=4^{1 / 3} \cdot 4^{-1 / 9} \cdot 4^{1 / 27} \ldots \infty and z=∑r=1∞(1+i)−rz=\sum\limits_{r=1}^{\infty}(1+i)^{-r}, where i=−1i=\sqrt{-1}, then arg⁡(x+yz)\arg (x+y z) is equal to
View answer and explanation
Correct option: B
−tan⁡−1(23)-\tan ^{-1}\left(\frac{\sqrt{2}}{3}\right)
Option B (-tan(-1)(sqrt(2)/3)) is correct because after simplifying x, y, and z, the argument of the resulting complex number is -tan(-1)(sqrt(2)/3).
38 Easy
38 of 612
Application +4 | -1
The maximum value of ∣z∣    where   z\left| z \right|\;\;\mathrm{where}\,\;z satisfies the condition ∣z+2z∣=2\left| {z + \frac{2}{z}} \right| = 2, is
View answer and explanation
Correct option: B
3+1\sqrt 3 + 1
Option B (3+1\sqrt 3 + 1) is correct because it represents the maximum possible value of the modulus of z that satisfies the given condition.
39 Easy
39 of 612
Analyzing +4 | -1
In the argand plane, if O,PO,P and QQ represent respectively the origin OO and the complex numbers zz and z+izz + iz respectively, then ∠OPQ\angle OPQ is
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi}{2}) is correct because the complex number z + iz is obtained by rotating z by 90 degrees counterclockwise in the Argand plane, making the angle OPQ a right angle.
40 Easy
40 of 612
Application +4 | -1
If 2z13z2\frac{{2{z_1}}}{{3{z_2}}} is purely imaginary number, then ∣z1−z2z1+z2∣4{\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} \right|^4} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because if 2z13z2\frac{2z_1}{3z_2} is purely imaginary, then by substituting z1z2=ki\frac{z_1}{z_2} = ki and simplifying the expression ∣z1−z2z1+z2∣4\left| \frac{z_1 - z_2}{z_1 + z_2} \right|^4, we get 1.
41 Easy
41 of 612
Analyzing +4 | -1
If 349(x+iy)=(32+i23)100,y∈N3^{49}(x+i y)=\left(\frac{3}{2}+\frac{i}{2} \sqrt{3}\right)^{100}, y \in \mathbf{N}, and x=kyx=k y, then value of kk is
View answer and explanation
Correct option: B
±22\pm 2 \sqrt{2}
Option B (±22\pm 2 \sqrt{2}) is correct because by equating the magnitudes and using the given condition x=ky and y being a natural number, we find that k can be ±22\pm 2 \sqrt{2}.
42 Easy
42 of 612
Analyzing +4 | -1
If x,  y,  zx,\;y,\;z are in GP and ax=by=cz{a^x} = {b^y} = {c^z}, then
View answer and explanation
Correct option: B
log⁡ba=log⁡cb{\log _b}a = {\log _c}b
Option B (log⁡ba=log⁡cb{\log _b}a = {\log _c}b) is correct because given that x, y, z are in GP and ax=by=cz{a^x} = {b^y} = {c^z}, by taking logarithms and using the properties of logarithms, we can derive the relationship log⁡ba=log⁡cb{\log _b}a = {\log _c}b.
43 Easy
43 of 612
Application +4 | -1
If zr=cos⁡(π2r)+isin⁡(π2r){z_r} = \cos \left( {\frac{\pi }{{{2^r}}}} \right) + i\sin \left( {\frac{\pi }{{{2^r}}}} \right), then z1.z2.z3…{z_1}.{z_2}.{z_3} \ldots upto ∞\infty equals
View answer and explanation
Correct option: D
−1- 1
Option D ( - 1) is correct because the infinite product of the given complex numbers simplifies to -1 using trigonometric identities.
44 Easy
44 of 612
Application +4 | -1
If 1,ω1, \omega and ω2\omega^2 are the cube roots of unity, then the value of (1+ω)3−(1+ω2)3(1+\omega)^3-\left(1+\omega^2\right)^3 is
View answer and explanation
Correct option: D
0
Option D (0) is correct because using the property 1 + ω = -ω² and 1 + ω² = -ω, the expression simplifies to 0.
45 Easy
45 of 612
Application +4 | -1
If log⁡42+log⁡44+log⁡4x+log⁡416=6{\log _4}2 + {\log _4}4 + {\log _4}x + {\log _4}16 = 6, then the value of xx is
View answer and explanation
Correct option: D
32
Option D (32) is correct because solving the logarithmic equation leads to x=32x = 32.

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46 Easy
46 of 612
Application +4 | -1
Let z1z_1 and z2z_2 be given by z1=(2+i52−i5)10 and z2=(2−i52+i5)10z_1=\left(\frac{2+i \sqrt{5}}{2-i \sqrt{5}}\right)^{10} \text { and } z_2=\left(\frac{2-i \sqrt{5}}{2+i \sqrt{5}}\right)^{10}
Then ∣z1+z2∣\left|z_1+z_2\right| is equal to
View answer and explanation
Correct option: A
2cos⁡(20cos⁡−123)2 \cos \left(20 \cos ^{-1} \frac{2}{3}\right)
Option C (2cos⁡(10cos⁡−123)2 \cos \left(10 \cos ^{-1} \frac{2}{3}\right)) is correct because by converting to polar form and using De Moivre's theorem, the expression simplifies to this result.
47 Easy
47 of 612
Understanding +4 | -1
Roots of x2+2=0x^2+2=0 are ±ni\pm \sqrt{n} i. The value of nn is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the roots of the equation are ±i2\pm i \sqrt{2}, therefore, n = 2.
48 Easy
48 of 612
Application +4 | -1
Let α,α2\alpha ,{\alpha ^2} be the roots of x2+x+1=0{x^2} + x + 1 = 0, then the equation whose roots are α31,  α62{\alpha ^{31}},\;{\alpha ^{62}} is
View answer and explanation
Correct option: C
x2+x+1=0{x^2} + x + 1 = 0
Option C (x2+x+1=0{x^2} + x + 1 = 0) is correct because the new roots α31{\alpha ^{31}} and α62{\alpha ^{62}} satisfy the same quadratic equation as α\alpha and α2{\alpha ^2}.
49 Easy
49 of 612
Application +4 | -1
If c+ic−i=a+ib\frac{c+i}{c-i}=a+i b, where a,b,ca, b, c are real, then a2+b2a^2+b^2 is equal to:
View answer and explanation
Correct option: A
1
Option A (1) is correct because when the given complex expression is simplified and the modulus is calculated, the result is 1.
50 Easy
50 of 612
Analyzing +4 | -1
If λ∈R\lambda \in {R} so that the origin and the non - real roots of the equation 2z2+2z+λ=02 z^2+2 z+\lambda=0 form the vertices of an equilateral triangle in the argand plane, then 1λ\frac{1}{\lambda} equal to
View answer and explanation
Correct option: A
03-Feb
The value of 1/λ is 3/2. This is found by using the properties of equilateral triangles and the relationship between the roots and coefficients of the quadratic equation.
51 Easy
51 of 612
Application +4 | -1
Find Re⁡(z)\operatorname{Re}(z), which satisfy ∣z∣=1|z|=1 and z−2z=1z-\frac{2}{z}=1
View answer and explanation
Correct option: B
−1-1
Option B (-1) is correct. By substituting z = x + yi and using the given conditions, we find that Re(z) = x = -1.
52 Easy
52 of 612
Analyzing +4 | -1
If z1,z2,z3{z_1},{z_2},{z_3} be vertices of an equilateral triangle occurring in the anticlockwise sense, then
View answer and explanation
Correct option: C
z1+ω  z2+ω2z3=0{z_1} + \omega \;{z_2} + {\omega ^2}{z_3} = 0
Option D (−2,2,4- 2,\mathrm{ }2,\mathrm{ }4) is correct because these are the solutions obtained by considering both positive and negative cases of the absolute value and verifying that they satisfy the original equation.
53 Easy
53 of 612
Application +4 | -1
For n=6  k,  k∈Z,(1−i32)n+(−1−i−32)nn = 6\;k,\;k \in Z,{\left( {\frac{{1 - i\sqrt 3 }}{2}} \right)^n} + {\left( {\frac{{ - 1 - i - \sqrt 3 }}{2}} \right)^n} has the value
View answer and explanation
Correct option: D
2
The correct answer is D (2) because substituting n=6kn = 6k into the expression and using the properties of complex cube roots of unity simplifies the expression to 2.
54 Easy
54 of 612
Application +4 | -1
The value of cos⁡30∘+isin⁡300cos⁡60∘−isin⁡600\frac{{\cos 30^\circ + i\sin {{30}^0}}}{{\cos 60^\circ - i\sin {{60}^0}}} is equal to
View answer and explanation
Correct option: A
ii
Option A (i) is correct because using Euler's formula and simplifying the expression results in the imaginary unit i.
55 Easy
55 of 612
Analyzing +4 | -1
All the points in the set S={α+iα−i:α∈R}(i=−1)S = \left\{ {\frac{{\alpha + i}}{{\alpha - i}}:\alpha \in R} \right\}(i = \sqrt { - 1} ) lie on a
View answer and explanation
Correct option: C
circle whose radius is 1.
Option C (circle whose radius is 1) is correct because the given expression represents a complex number whose magnitude is always 1, thus lying on a circle of radius 1 centered at the origin.
56 Easy
56 of 612
Analyzing +4 | -1
If aa and bb are real numbers between 0 and 1 such that the points z1=a+i,z2=1+biz_1=a+i, z_2=1+b i and z3=0z_3=0 form an equilateral triangle, then
View answer and explanation
Correct option: A
a=b=2−3a=b=2-\sqrt{3} should look like this
Option A (a=b=2−3a=b=2-\sqrt{3}) is correct because it satisfies the equilateral triangle condition and the constraint that a and b are between 0 and 1.
57 Easy
57 of 612
Application +4 | -1
If z1,  z2{z_1},\;{z_2} and z3{z_3} are any three complex numbers, then the fourth vertex of the parallelogram whose three vertices are z1,  z2{z_1},\;{z_2} and z3{z_3} taken in order is
View answer and explanation
Correct option: A
z1−z2+z3{z_1} - {z_2} + {z_3}
Option A (z1−z2+z3{z_1} - {z_2} + {z_3}) is correct because in a parallelogram, the midpoints of the diagonals coincide, allowing us to find the fourth vertex.
58 Easy
58 of 612
Application +4 | -1
If ∣z−3−2i∣=∣z+2i∣|z-3-2 i|=|z+2 i|, where zz is a complex number, then minimum value of ∣z∣|z| will be
View answer and explanation
Correct option: D
910\frac{9}{10}
Option D (910\frac{9}{10}) is correct because the equation represents a straight line, and the minimum distance from the origin to the line is 910\frac{9}{10}.
59 Easy
59 of 612
Application +4 | -1
kk non-overlapping regular polygons of p1,p2,…,pkp_1, p_2, \ldots, p_k sides have one common vertex so that no gap is left at that vertex. Then 1p1+1p2+⋯+1pk\frac{1}{p_1}+\frac{1}{p_2}+\cdots+\frac{1}{p_k} is equal to
View answer and explanation
Correct option: B
k2−1\frac{k}{2}-1
The value of (1+isin⁡π8+cos⁡π81−isin⁡π8+cos⁡π8)8\left(\frac{{1+i \sin \frac{{\pi}}{{8}}+\cos \frac{{\pi}}{{8}}}}{{1-i \sin \frac{{\pi}}{{8}}+\cos \frac{{\pi}}{{8}}}}\right)^8 is -1 because it simplifies to -1 using trigonometric identities and De Moivre's theorem.
60 Easy
60 of 612
Analyzing +4 | -1
−1−1−1−...∞  \sqrt { - 1 - \sqrt {1 - \sqrt {1 - ...\infty } } \;} is equal to
View answer and explanation
Correct option: C
ω2{\omega ^2}
Option C (ω2{\omega ^2}) is correct because solving the given equation leads to complex roots, which are cube roots of unity.
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