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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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271 Easy
271 of 612
Understanding +4 | -1
If z1,z2{z_1},{z_2} are any two complex numbers, then
View answer and explanation
Correct option: C
∣z1+z2∣≤∣z1∣+∣z2∣\left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|
Option C is correct because the triangle inequality states that the magnitude of the sum of two complex numbers is less than or equal to the sum of their magnitudes.
272 Easy
272 of 612
Application +4 | -1
Suppose arg⁡(z)=−5π/13\arg (z)=-5 \pi / 13, then arg⁡(z+zˉ1+zzˉ)\arg \left(\frac{z+\bar{z}}{1+z \bar{z}}\right) is
View answer and explanation
Correct option: C
π\pi
Option D (0) is correct because (z + conjugate(z))/(1 + z*conjugate(z)) is a real number, and the argument of any positive real number is 0.
273 Easy
273 of 612
Analyzing +4 | -1
If z≠1z \neq 1 and z2z−1\frac{z^2}{z-1} is real, the point represented by the complex number zz lies
View answer and explanation
Correct option: D
either on the real axis or on a circle passing through the origin
Option D (either on the real axis or on a circle passing through the origin) is correct because the imaginary part of the expression being real leads to the equation of a circle passing through the origin or the real axis.
274 Easy
274 of 612
Application +4 | -1
The equation z2=zˉ{z^2} = \bar z has
View answer and explanation
Correct option: C
Four solutions
Option C (Four solutions) is correct because the equation z2=zˉ{z^2} = \bar z has four solutions: three from z3=1{z^3} = 1 and one at z = 0.
275 Easy
275 of 612
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 derived by substituting x=9x=9 into the factored form of xn−1x^n - 1 divided by x−1x-1, where the roots of unity are used.
276 Easy
276 of 612
Application +4 | -1
The value of Re⁡[(1+i)23−i]\operatorname{Re}\left[\frac{(1+i)^2}{3-i}\right] is equal to
View answer and explanation
Correct option: A
−15-\frac{1}{5}
Option A (-15\frac{{1}}{{5}}) is correct because the real part of the simplified expression is -15\frac{{1}}{{5}}.
277 Easy
277 of 612
Application +4 | -1
If the amplitude of z−2−3iz - 2 - 3i is π4\frac{\pi }{4}, then the locus of z=x+iyz = x + iy, is
View answer and explanation
Correct option: D
x−y+1=0x - y + 1 = 0
Option D (x−y+1=0x - y + 1 = 0) is correct because it represents the locus of z, derived from the given amplitude condition by equating the tangent of the angle to 1.
278 Easy
278 of 612
Application +4 | -1
If z=(3+i2)5+(3−i2)5z=\left(\frac{\sqrt{3}+i}{2}\right)^5+\left(\frac{\sqrt{3}-i}{2}\right)^5, then
View answer and explanation
Correct option: B
Im⁡(z)=0\operatorname{Im}(z)=0
Option B (Im⁡(z)=0\operatorname{Im}(z)=0) is correct because the imaginary part of the complex number zz is zero, making it a real number.
279 Easy
279 of 612
Application +4 | -1
If z1,z2,z3{z_1},{z_2},{z_3} be three complex numbers such that ∣z1+1∣≤1,∣z2+2∣≤2  and  ∣z3+4∣≤4\left| {{z_1} + 1} \right| \le 1,\left| {{z_2} + 2} \right| \le 2\;\mathrm{and}\;\left| {{z_3} + 4} \right| \le 4 then the maximum value of ∣z1∣+∣z2∣+∣z3∣\left| {{z_1}} \right| + \left| {{z_2}} \right| + \left| {{z_3}} \right| is
View answer and explanation
Correct option: D
14
Option D (14) is correct because it represents the maximum possible value of the sum of the magnitudes of the complex numbers, which occurs when each complex number lies on the circle's circumference farthest from the origin.
280 Easy
280 of 612
Application +4 | -1
α1,α2,α3,…,α100\alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_{100} are all the 100th roots of unity. The numerical value of ∑1≤j≤j≤100(αjαj)5\sum_{1 \leq j \leq j \leq 100}\left(\alpha_j \alpha_j\right)^5 is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the sum of the fifth powers of the 100th roots of unity is zero.
281 Easy
281 of 612
Application +4 | -1
i57+1i125i^{57}+\frac{1}{i^{125}}, when simplified has the value
View answer and explanation
Correct option: A
0
Option A (0) is correct because after simplification, the expression equals 0.
282 Easy
282 of 612
Application +4 | -1
The value of (1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9)3\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^{3} is
View answer and explanation
Correct option: D
−12(3−i)-\frac{1}{2}(\sqrt{3}-i)
Option D (-12(3−i)\frac{1}{2}(\sqrt{3}-i)) is correct because it is the simplified form of the given complex expression after applying Euler's formula.
283 Easy
283 of 612
Understanding +4 | -1
Consider the following statements.

L. Representation of z=x+iy{z}={x}+{iy} in terms of r{r} and θ\theta is called polar form of the complex number.

II. arg⁡(z1z2)=arg⁡(z1)−arg⁡(z2)\arg \left(\frac{z_1}{z_2}\right)=\arg \left(z_1\right)-\arg \left(z_2\right)

Choose the correct option.
View answer and explanation
Correct option: D
Both I and II are correct
Option D is correct because both statements are correct. Statement I defines the polar form, and Statement II describes the argument of a quotient of complex numbers.
284 Easy
284 of 612
Analyzing +4 | -1
If ∣z−2−i∣=∣z∣∣sin⁡(π4−arg⁡z)∣|z-2-i|=|z|\left|\sin \left(\frac{\pi}{4}-\arg z\right)\right|, where i=−1i=\sqrt{-1}, then locus of zz, is
View answer and explanation
Correct option: C
parabola
Option C (parabola) is correct because the given equation represents a parabola when converted to Cartesian coordinates.
285 Easy
285 of 612
Application +4 | -1
Let α\alpha and β\beta be the roots of x2−2xcos⁡∅+1=0{x^2} - 2x\cos \emptyset + 1 = 0. Then the equation whose roots are αn,  βn{\alpha ^n},\;{\beta ^n} is
View answer and explanation
Correct option: B
x2−2xcos⁡n∅+1=0{x^2} - 2x\cos n\emptyset + 1 = 0
Option B (x2−2xcos⁡n∅+1=0{x^2} - 2x\cos n\emptyset + 1 = 0) is correct because the roots αn{\alpha ^n} and βn{\beta ^n} satisfy this equation, derived using De Moivre's theorem.
286 Easy
286 of 612
Application +4 | -1
If zz is a complex number such that ∣z∣≥2|z| \geq 2, then the minimum value of ∣z+(12)∣\left|z+\left(\frac{1}{2}\right)\right|, is
View answer and explanation
Correct option: D
lies in the interval (1,2)(1,2)
The correct answer is D (lies in the interval (1,2)) because the minimum value of ∣z+(1/2)∣|z + (1/2)| occurs when zz is on the circle ∣z∣=2|z| = 2 and closest to −1/2-1/2, resulting in a value of 3/2, which lies in the interval (1,2).
287 Easy
287 of 612
Analyzing +4 | -1
Let Z1Z_{1} and Z2Z_{2} be two roots of the equation Z2+aZ+b=0,ZZ^{2}+a Z+b=0, Z being complex. Further, assume that the origin, Z1Z_{1} and Z2Z_{2} form an equilateral triangle. Then
View answer and explanation
Correct option: D
a2=3ba^{2}=3 b
Option D (a2=3ba^{2}=3 b) is correct because it is the relationship derived from the equilateral triangle condition and Vieta's formulas.
288 Easy
288 of 612
Analyzing +4 | -1
If zˉ=3i+25z+3i\bar{z}=3 i+\frac{25}{z+3 i}, then ∣z∣|z| cannot exceed
View answer and explanation
Correct option: B
8
Option B (8) is correct because it represents the maximum possible value of |z| derived from the given equation by using the triangle inequality.
289 Easy
289 of 612
Application +4 | -1
Argument of the complex number (−1−3i2+i)\left( {\frac{{ - 1 - 3i}}{{2 + i}}} \right) is
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Correct option: C
2250{225^0}
Option C (2250{225^0}) is correct because the argument of the simplified complex number -1-i lies in the third quadrant.
290 Easy
290 of 612
Analyzing +4 | -1
If z2+z+1=0{z^2} + z + 1 = 0, where zz is a complex number, then the value of (z+1z)2+(z2+1z2)2+(z3+1z3)+...+(z6+1z6)2{\left( {z + \frac{1}{z}} \right)^2} + {\left( {{z^2} + \frac{1}{{{z^2}}}} \right)^2} + \left( {{z^3} + \frac{1}{{{z^3}}}} \right) + ... + {\left( {{z^6} + \frac{1}{{{z^6}}}} \right)^2}
View answer and explanation
Correct option: B
12
Option B is correct because by substituting z=ωz = \omega or z=ω2z = \omega^2, the expression simplifies to 12, where ω\omega is a complex cube root of unity.
291 Easy
291 of 612
Analyzing +4 | -1
Let α\alpha and β\beta be two distinct complex numbers, such that ∣α∣=∣β∣|\alpha|=|\beta|. If real part of α\alpha is positive and imaginary part of β\beta is negative, then the complex number (α+β)/(α−β)(\alpha+\beta) /(\alpha-\beta) may be
View answer and explanation
Correct option: D
purely imaginary
Option D (purely imaginary) is correct because the expression simplifies to a purely imaginary number using Euler's formula and trigonometric identities.
292 Easy
292 of 612
Application +4 | -1
If ∣z∣≥5|z| \geq 5, then least value of ∣z−1z∣\left|z-\frac{1}{z}\right| is
View answer and explanation
Correct option: B
24/524 / 5
Option B (24/524 / 5) is correct because the least value of the expression is found by using the triangle inequality and substituting the minimum value of |z|.
293 Easy
293 of 612
Application +4 | -1
The imaginary part of (1+i)2i(2i−1)\frac{{{{\left( {1 + i} \right)}^2}}}{{i\left( {2i - 1} \right)}} is
View answer and explanation
Correct option: D
−(4/5)- \left( {4/5} \right)
Option D (−(4/5)- \left( {4/5} \right)) is correct because it is the imaginary part of the simplified complex number.
294 Easy
294 of 612
Analyzing +4 | -1
If a,b,c,a1,b1a, b, c, a_1, b_1 and c1c_1 are non-zero complex numbers satisfying aa1+bb1+cc1=1+i\frac{a}{a_1}+\frac{b}{b_1}+\frac{c}{c_1}=1+i and a1a+b1b+c1c=0\frac{a_1}{a}+\frac{b_1}{b}+\frac{c_1}{c}=0, where i=−1i=\sqrt{-1}, the value of a2a12+b2b12+c2c12\frac{a^2}{a_1^2}+\frac{b^2}{b_1^2}+\frac{c^2}{c_1^2} is
View answer and explanation
Correct option: A
2i2 i
Option A (2i2i) is correct because by squaring the first equation and using the second equation, the value of the expression can be found.
295 Easy
295 of 612
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} equals
View answer and explanation
Correct option: A
4(a2−b2)4\left(a^2-b^2\right)
Option A (4(a2−b2)4(a^2-b^2)) is correct because it results from equating real and imaginary parts of the complex equation and simplifying the expression.
296 Easy
296 of 612
Application +4 | -1
If π2<α<3π2,\frac{\pi }{2} < \alpha < \frac{{3\pi }}{2}, the modulus and argument form of (1+cos⁡2  α)+isin⁡2  α\left( {1 + \cos 2\;\mathrm{\alpha }} \right) + i\sin 2\;\mathrm{\alpha } is
View answer and explanation
Correct option: A
−2cos⁡α{cos⁡(π+α)+isin⁡(π+α)}- 2\cos \mathrm{\alpha }\left\{ {\cos \left( {\pi + \alpha } \right) + i\sin \left( {\pi + \alpha } \right)} \right\}
Option A (-2\cos\mathrm{\alpha}\left{\cos \left( {\pi + \alpha } \right) + i\sin \left( {\pi + \alpha } \right)} \right}-2\cos\mathrm{\alpha}\left{\cos \left( {\pi + \alpha } \right) + i\sin \left( {\pi + \alpha } \right)} \right}) is correct because it correctly represents the modulus and argument of the given complex number in the specified range for α\alpha.
297 Easy
297 of 612
Application +4 | -1
The complex number z1,z2,z3z_1, z_2, z_3 are the vertices of an equilateral triangle. If z0z_0 is the circumcentre of the triangle, then z12+z22+z32z_1^2+z_2^2+z_3^2 is equal to
View answer and explanation
Correct option: B
3z023 z_0^2
Option B (3z023 z_0^2) is correct because in an equilateral triangle, the circumcenter coincides with the centroid, and using the properties of equilateral triangles, z12+z22+z32=3z02z_1^2+z_2^2+z_3^2 = 3z_0^2.
298 Easy
298 of 612
Application +4 | -1
If 1,ω,ω21, \omega, \omega^2 are the three cube roots of unity, then (1−ω+ω2)(1−ω2+ω4)(1−ω4+ω8)…\left(1-\omega+\omega^2\right)\left(1-\omega^2+\omega^4\right)\left(1-\omega^4+\omega^8\right) \ldots to 2n2 n factors ==
View answer and explanation
Correct option: B
22n2^{2 n}
Option B (22n2^{2n}) is correct because the expression simplifies to 22n2^{2n} using the properties of cube roots of unity.
299 Easy
299 of 612
Analyzing +4 | -1
If the equation a∣z∣2+α‾z+αz‾‾+d=0a{\left| z \right|^2} + \overline {\overline \alpha z + \alpha \overline z } + d = 0 represents a circle, where a, da,\,d are real constants, then which of the following condition is correct?
View answer and explanation
Correct option: B
∣α∣2−ad>0  and  a∈R−{0}{\left| \alpha \right|^2} - ad > 0\,\,\mathrm{and}\,\,a \in R - \{ 0\}
Option B (∣α∣2−ad>0, mathrmand, a∈R−0{\left| \alpha \right|^2} - ad > 0\\,\,\\mathrm{and}\\,\,a \in R - \\{ 0\\}) is correct because it represents the condition for the given equation to represent a circle.
300 Easy
300 of 612
Application +4 | -1
If 32+cos⁡θ+isin⁡θ=a+ib\frac{3}{2+\cos \theta+i \sin \theta}=a+i b, then (a−2)2+b2(a-2)^2+b^2 equals
View answer and explanation
Correct option: B
1
Option B (1) is correct because after manipulating the given equation and substituting into the expression, the result simplifies to 1.
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