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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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271 Easy
271 of 979
Understanding +4 | -1
The equation ∣z+1−i∣=∣z+i−1∣|z+1-i|=|z+i-1| represents
View answer and explanation
Correct option: A
a straight line
Option A (a straight line) is correct because the equation represents the locus of points equidistant from two fixed points, which is a straight line (perpendicular bisector).
272 Easy
272 of 979
Application +4 | -1
Let ω\omega be a complex cube root of unity. If the equation ∣z−ω∣2+∣z−ω2∣2=λ{\left| {z - \omega } \right|^2} + {\left| {z - {\omega ^2}} \right|^2} = \lambda represents a circle with points representing ω\omega and ω2{\omega ^2} as the end points of a diameter, then λ=\lambda =
View answer and explanation
Correct option: B
3
The correct answer is B (3) because the diameter squared is the distance between the cube roots of unity.
273 Medium
273 of 979
Application +4 | -1
If w=zz−13iw = \frac{z}{{z - \frac{1}{3}i}} and ∣w∣=1\left| w \right| = 1, then zz lies on
View answer and explanation
Correct option: B
A straight line
A straight line. Option B is correct because the condition |w| = 1 leads to a linear equation in terms of the real and imaginary parts of z.
274 Easy
274 of 979
Application +4 | -1
If ω\omega is fifth root of unity, then log⁡2∣1+ω+ω2+ω3−ω−1∣\log _2\left|1+\omega+\omega^2+\omega^3-\omega^{-1}\right| is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because simplifying the expression using the properties of the fifth root of unity results in a value of 1.
275 Easy
275 of 979
Analyzing +4 | -1
Let z=cos⁡θ+isin⁡θz=\cos \theta+i \sin \theta. Then the value of ∑m=115Im⁡(z2m−1)\sum\limits_{{m=1}}^{15} \operatorname{Im}\left(z^{2 m-1}\right) at θ=2∘\theta=2^{\circ} is
View answer and explanation
Correct option: D
14sin⁡2∘\frac{1}{4 \sin 2^{\circ}}
Option D (14sin⁡2∘\frac{1}{4 \sin 2^{\circ}}) is correct because by using the formula for the sum of a geometric series and trigonometric identities, the given sum simplifies to 14sin⁡2∘\frac{1}{4 \sin 2^{\circ}}.
276 Medium
276 of 979
Application +4 | -1
Let z∈Cz \in \mathbf{C} be such that Re⁡(z2)=0\operatorname{Re}\left(z^2\right)=0, then
View answer and explanation
Correct option: B
∣Re⁡(z)∣=∣Im⁡(z)∣|\operatorname{Re}(z)|=|\operatorname{Im}(z)|
Option B (∣Re⁡(z)∣=∣Im⁡(z)∣|\operatorname{Re}(z)|=|\operatorname{Im}(z)|) is correct because if Re⁡(z2)=0\operatorname{Re}(z^2)=0, then the real and imaginary parts of z must have equal magnitudes.
277 Easy
277 of 979
Application +4 | -1
The least positive integer nn for which (1+i1−i)n{\left( {\frac{{1 + i}}{{1 - i}}} \right)^n} is real, is
View answer and explanation
Correct option: A
2
The correct answer is A (2) because it is the smallest positive integer nn for which (1+i1−i)n{\left( {\frac{{1 + i}}{{1 - i}}} \right)^n} is a real number.
278 Medium
278 of 979
Application +4 | -1
If z=ilog⁡(2−3),  thencos⁡z=z = i\log \left( {2 - \sqrt 3 } \right),\;\mathrm{then}\cos z =
View answer and explanation
Correct option: D
2
Option B (−15-15) is correct because by substituting x = 3 + i into the given expression and simplifying, we find that the result is -15.
279 Medium
279 of 979
Application +4 | -1
Let z=x+iyz=x+i y, where xx and yy are real. The points (x,y)(x, y) in the plane, for which (z+i)/(z−i)(z+i) /(z-i) is purely imaginary, lie on
View answer and explanation
Correct option: B
a circle
Option B (a circle) is correct because the condition for the expression to be purely imaginary leads to the equation x² + y² = 1, which represents a circle.
280 Easy
280 of 979
Application +4 | -1
The vector z=−4+5  iz = - 4 + 5\;i is turned counterclockwise through an angle of 180∘180^\circ and stretched 1121\frac{1}{2} times. The complex number corresponding to newly obtained vector is
View answer and explanation
Correct option: A
6−152i6 - \frac{{15}}{2}i
Option A (6−152i6 - \frac{{15}}{2}i) is correct because it is the result of rotating the vector counterclockwise by 180 degrees and stretching it by a factor of 1.5.
281 Easy
281 of 979
Application +4 | -1
If ω=cos⁡πn+isin⁡πn\omega=\cos \frac{\pi}{n}+i \sin \frac{\pi}{n}, then value of 1+ω+ω2+…+ωn−11+\omega+\omega^2+\ldots+\omega^{n-1} is
View answer and explanation
Correct option: A
1+icot⁡(π2π)1+i \cot \left(\frac{\pi}{2 \pi}\right)
Option A (1+icot⁡(π2π)1+i \cot (\frac{{\pi}}{{2 \pi}})) is correct because it is derived from summing the geometric series and simplifying using trigonometric identities.
282 Easy
282 of 979
Analyzing +4 | -1
Suppose z∈Cz \in \mathbf{C}, and z∉Rz \notin \mathbf{R}. If w=1−z+z21+z+z2w=\frac{1-z+z^2}{1+z+z^2} is a real number, then ∣z∣|z| equals
View answer and explanation
Correct option: A
1
Option A (1) is correct because the condition that ww is real implies that z+1zz + \frac{{1}}{{z}} is real, which leads to ∣z∣=1|z| = 1 since zz is not real.
283 Medium
283 of 979
Application +4 | -1
If (x+iy)1/3=a+ib(x+i y)^{1 / 3}=a+i b, then xa+yb\frac{x}{a}+\frac{y}{b} is equal to
View answer and explanation
Correct option: B
4(a2−b2)4\left(a^2-b^2\right)
Option B (4(a2−b2)4(a^2-b^2)) is correct because cubing both sides of the given equation and equating real and imaginary parts allows us to express x/a + y/b in terms of a2a^2 and b2b^2.
284 Easy
284 of 979
Analyzing +4 | -1
If z1=1+2i,    z2=2+3i,    z3=3+4i,{z_1} = 1 + 2i,\;\;{z_2} = 2 + 3i,\;\;{z_3} = 3 + 4i, then z1,  z2,  z3{z_1},\;{z_2},\;{z_3} represents the vertices of a/an
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the given complex numbers represent collinear points, not vertices of any triangle.
285 Medium
285 of 979
Application +4 | -1
If 1,ω,ω2,….ωn−11, \omega, \omega^2, \ldots . \omega^{n-1} are nn, nth roots of unity, then the value of (9−ω)(9−ω2)….(9−ωn−1)(9-\omega)\left(9-\omega^2\right) \ldots .\left(9-\omega^{n-1}\right) will be
View answer and explanation
Correct option: C
9n−18\frac{9^n-1}{8}
Option C (9n−18\frac{9^n-1}{8}) is correct because it is obtained by substituting x=9 into the factored form of xn−1x^n - 1.

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286 Medium
286 of 979
Analyzing +4 | -1
For all z∈Cz \in C on the curve C1:∣z∣=4C_{1}:|z|=4, let the locus of the point z+1zz+\frac{1}{z} be the curve C2C_{2}. Then:
View answer and explanation
Correct option: C
The curves C1C_{1} and C2C_{2} intersect at 4 points
The curves C1C_1 and C2C_2 intersect at 4 points because C1C_1 is a circle and C2C_2 is an ellipse where the radius of the circle is smaller than the major axis of the ellipse.
287 Easy
287 of 979
Analyzing +4 | -1
If 8iz3+12z2−18z+27i=08 i z^3+12 z^2-18 z+27 i=0, then
View answer and explanation
Correct option: A
∣z∣=32|z|=\frac{3}{2}
Option A (∣z∣=32|z|=\frac{{3}}{{2}}) is correct because by factoring the given equation, we find that ∣z∣=32|z| = \frac{{3}}{{2}} is a solution.
288 Medium
288 of 979
Application +4 | -1
If a>0a > 0 and the equation ∣z−a2∣+∣z−2a∣=3\left| {z - {a^2}} \right| + \left| {z - 2a} \right| = 3 represents an ellipse, then aa belongs to the interval
View answer and explanation
Correct option: C
(0,  3)\left( {0,\;3} \right)
The correct answer is C ((0, 3)) because the condition for the equation to represent an ellipse requires the distance between the foci to be less than the length of the major axis, leading to the inequality ∣a2−2a∣<3|a^2 - 2a| < 3.
289 Medium
289 of 979
Application +4 | -1
If α\alpha and β\beta are different complex numbers with ∣β∣=1\left| \beta \right| = 1, then ∣β−α1−αˉβ∣\left| {\frac{{\beta - \alpha }}{{1 - \bar \alpha \beta }}} \right| is
View answer and explanation
Correct option: D
1
Option D (1) is correct because the expression simplifies to 1 using the property that |β| = 1.
290 Medium
290 of 979
Application +4 | -1
If the points A(z),B(−z)\mathrm{A}(\mathrm{z}), \mathrm{B}(-\mathrm{z}) and C(z−1)\mathrm{C}(\mathrm{z}-1) form an equilateral triangle then value of z\mathrm{z} is
View answer and explanation
Correct option: B
1±3i4\frac{1 \pm \sqrt{3} i}{4}
Option B (1±3i4\frac{{1 \pm \sqrt{{3}} i}}{{4}}) is correct because it is the solution for z when the points A(z), B(-z), and C(z-1) form an equilateral triangle.
291 Easy
291 of 979
Application +4 | -1
If z1=2+3iz_1=2+3 i and z2=3+2iz_2=3+2 i, then z1+z2z_1+z_2 equals to a+ai.a+ ai. Value of &#39;aa&#39; is equal to
View answer and explanation
Correct option: A
5
Option A (5) is correct because adding the two complex numbers z1 and z2 results in 5 + 5i, so the value of 'a' is 5.
292 Easy
292 of 979
Application +4 | -1
If zz lies on the circle ∣z−1∣=1,\left| {z - 1} \right| = 1, z−2z\frac{{z - 2}}{z} is
View answer and explanation
Correct option: B
Purely imaginary
Option B (Purely imaginary) is correct because the expression z−2z\frac{z-2}{z} simplifies to itan⁡(θ2)i \tan(\frac{\theta}{2}), which is purely imaginary.
293 Easy
293 of 979
Application +4 | -1
If (a1+ib1)(a2+ib2)⋯(an+ibn)=A+iB\left(a_1+i b_1\right)\left(a_2+i b_2\right) \cdots\left(a_n+i b_n\right)=A+i B, then (a12+b12)(a22+b22)⋯(an2+bn2)\left(a_1^2+b_1^2\right)\left(a_2^2+b_2^2\right) \cdots\left(a_n^2+b_n^2\right) is equal to
View answer and explanation
Correct option: B
A2+B2A^2+B^2
Option B (A2+B2A^2+B^2) is correct because it represents the product of the magnitudes of the complex numbers.
294 Medium
294 of 979
Application +4 | -1
If C2+S2=1C^2+S^2=1, then 1+C+iS1+C−iS\frac{1+C+i S}{1+C-i S} is equal to
View answer and explanation
Correct option: A
C+iSC+i S
Option A (C+iSC+i S) is correct because the expression simplifies to C+iSC+iS using trigonometric identities.
295 Medium
295 of 979
Application +4 | -1
3+2isin⁡θ1−2isin⁡θ\frac{{3 + 2i\sin \theta }}{{1 - 2i\sin \theta }} will be purely imaginary, if θ\theta is equal to
View answer and explanation
Correct option: C
nπ±π3n\pi \pm \frac{\pi }{3}
Option C (nπ±π3n\pi \pm \frac{\pi }{3}) is correct because setting the real part of the expression to zero yields this general solution for θ\theta.
296 Easy
296 of 979
Application +4 | -1
If −π<arg⁡(z)<−π2- \pi < \arg \left( z \right) < - \frac{\pi }{2} then arg⁡(zˉ)−arg⁡(−zˉ)\arg \left( {\bar z} \right) - \arg \left( { - \bar z} \right) is
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because given the range of arg(z), the expression simplifies to π\pi using properties of arguments of complex numbers.
297 Medium
297 of 979
Application +4 | -1
∣(1+i)(2+i3+i)∣\left|(1+i)\left(\frac{2+i}{3+i}\right)\right| is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because the modulus of a product is the product of the moduli, and the modulus of a quotient is the quotient of the moduli, leading to the answer 1.
298 Easy
298 of 979
Application +4 | -1
If zz is purely imaginary and Im⁡(z)<0\operatorname{Im}(z) < 0, then arg⁡(izˉ)+arg⁡(z)\arg (i \bar{z})+\arg (z) is equal to
View answer and explanation
Correct option: C
π/2\pi / 2
Option C (π/2\pi / 2) is correct because the arguments of the complex numbers add up to this value, given the conditions on z.
299 Easy
299 of 979
Understanding +4 | -1
For any complex number zz, the minimum value of ∣z∣+∣z−1∣\left| z \right| + \left| {z - 1} \right| is
View answer and explanation
Correct option: B
1
Option B (1) is correct because by the triangle inequality, ∣z∣+∣z−1∣≥∣z−(z−1)∣=∣1∣=1|z| + |z-1| \ge |z - (z-1)| = |1| = 1.
300 Easy
300 of 979
Application +4 | -1
If ω(≠1)\omega(\neq 1) is a cube root of unity and (1+ω2)11=a+bω+cω2\left(1+\omega^2\right)^{11}=a+b \omega+c \omega^2, then (a,b,c)(a, b, c) equals
View answer and explanation
Correct option: A
(1,1,0)(1,1,0)
Option A (1,1,0) is correct because simplifying (1+ω^2)11 using the property 1+ω+ω^2=0 leads to the expression 1+ω, which corresponds to a=1, b=1, and c=0.
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