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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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661 Medium
661 of 979
Application +4 | -1
cos⁡(ilog⁡a−iba+ib)\cos \left(i \log \frac{a-i b}{a+i b}\right) is equal to
View answer and explanation
Correct option: B
a2−b2a2+b2\frac{a^2-b^2}{a^2+b^2}
Option B (a2−b2a2+b2\frac{a^2-b^2}{a^2+b^2}) is correct because it is the simplified form of cos⁡(ilog⁡a−iba+ib)\cos(i \log \frac{a-ib}{a+ib}) using complex number properties.
662 Medium
662 of 979
Application +4 | -1
(cos⁡2θ+isin⁡2θ)−5(cos⁡3θ−isin⁡3θ)6(\cos 2 \theta+i \sin 2 \theta)^{-5}(\cos 3 \theta-i \sin 3 \theta)^6 (sin⁡θ−icos⁡θ)3(\sin \theta-i \cos \theta)^3 is equal to
View answer and explanation
Correct option: C
sin⁡25θ+icos⁡25θ\sin 25 \theta+i \cos 25 \theta
Option C (sin⁡25θ+icos⁡25θ\sin 25 \theta + i \cos 25 \theta) is correct because after applying De Moivre's theorem and simplifying, the expression equals sin⁡25θ+icos⁡25θ\sin 25 \theta + i \cos 25 \theta.
663 Medium
663 of 979
Analyzing +4 | -1
If for the complex numbers z1z_1 and z2z_2, ∣1−zˉ1z2∣2−∣z1−z2∣2=k(1−∣z1∣2)(1−∣z2∣2)\left|1-\bar{z}_1 z_2\right|^2-\left|z_1-z_2\right|^2=k\left(1-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right), then kk is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because by expanding and simplifying the given expression, it reduces to (1−∣z1∣2)(1−∣z2∣2)(1-|z_1|^2)(1-|z_2|^2), which implies k=1k=1.
664 Easy
664 of 979
Analyzing +4 | -1
The solution set of x2+2≤3x≤2x2−5,{x^2} + 2 \le 3x \le 2{x^2} - 5, is
View answer and explanation
Correct option: A
ϕ\phi
Option A (ϕ\phi) is correct because there is no value of x that satisfies both inequalities derived from the given compound inequality.
665 Easy
665 of 979
Application +4 | -1
If (5+3i)33=249  z{\left( {\sqrt 5 + \sqrt 3 i} \right)^{33}} = {2^{49}}\;z, then modulus of the complex number zz is equal to
View answer and explanation
Correct option: B
2\sqrt 2
Option B (2\sqrt{2}) is correct because the modulus of z is calculated using De Moivre's theorem and properties of complex numbers.
666 Easy
666 of 979
Application +4 | -1
The complex number z=1+i\mathrm{z}=1+\mathrm{i} is rotated through an angle 3π2\frac{3 \pi}{2} in anticlockwise direction about the origin and stretched by additional 2\sqrt{2} unit, then the new complex number is:
View answer and explanation
Correct option: D
2(1−i)2(1-i)
Option D (2(1−i)2(1-i)) is correct because it represents the complex number after rotation and stretching as described in the problem.
667 Easy
667 of 979
Application +4 | -1
If z=5+t+i25−t2z=5+t+i \sqrt{25-t^2}, (−5≤t≤5)(-5 \leq t \leq 5), then locus of zz is a curve which passes through
View answer and explanation
Correct option: C
2+4i2+4 i
Option C (2+4i2+4i) is correct because substituting x=2x=2 and y=4y=4 into the locus equation (x−5)2+y2=25(x-5)^2 + y^2 = 25 satisfies the equation.
668 Easy
668 of 979
Application +4 | -1
The point (4, 1) undergoes the following three transformations successively (i) Refletion about the line y=xy = x
(ii) Translation through a distance of 2 unit along the positive direction of xx-axis
(iii) Rotation through an angle of π4\frac{\pi }{4} about the origin in the anti-clockwise direction

The final position of the point is
View answer and explanation
Correct option: B
(−12,  72)\left( { - \frac{1}{{\sqrt 2 }},\;\frac{7}{{\sqrt 2 }}} \right)
Option B (−12,;72)\left( { - \frac{{1}}{{\sqrt 2 }},\\;\frac{{7}}{{\sqrt 2 }}} \right) is correct because it represents the final position of the point after reflection, translation, and rotation.
669 Easy
669 of 979
Application +4 | -1
The value of [(cos⁡20∘+isin⁡20∘)(cos⁡75∘+isin⁡75∘)(cos⁡10∘+isin⁡10∘)]sin⁡15∘−icos⁡15∘\frac{{\left[ {(\cos 20^\circ + i\sin 20^\circ )\left( {\cos 75^\circ + i\sin 75^\circ } \right)\left( {\cos 10^\circ + i\sin 10^\circ } \right)} \right]}}{{\sin 15^\circ - i\cos 15^\circ }} is
View answer and explanation
Correct option: B
−1-1
Option B (-1) is correct because simplifying the expression using Euler's formula and trigonometric identities leads to -1.
670 Easy
670 of 979
Analyzing +4 | -1
If cos⁡α+cos⁡β+cos⁡γ=sin⁡α+sin⁡β+sin⁡γ=0,  thencos⁡3α+cos⁡3β+cos⁡3γ  \cos \mathrm{\alpha } + \cos \mathrm{\beta } + \cos \mathrm{\gamma } = \sin \mathrm{\alpha } + \sin \mathrm{\beta } + \sin \mathrm{\gamma } = 0,\mathrm{\;then}\cos 3\mathrm{\alpha } + \cos 3\mathrm{\beta } + \cos 3\mathrm{\gamma \;} equals
View answer and explanation
Correct option: C
3cos⁡(α+β+γ)3\cos \left( {\mathrm{\alpha } + \mathrm{\beta } + \mathrm{\gamma }} \right)
Option C (3cos⁡(α+β+γ)3\cos({\alpha} + {\beta} + {\gamma})) is correct because using complex number representation and the condition a+b+c=0a+b+c=0, we can derive the given trigonometric identity.
671 Medium
671 of 979
Application +4 | -1
If ∣z−i∣=1\left| {z - i} \right| = 1 and arg⁡(z)=θ,\arg \left( z \right) = \mathrm{\theta }, where 0<θ<π20 < \mathrm{\theta } < \frac{\pi }{2}, then cot⁡θ−2z\cot \mathrm{\theta } - \frac{2}{z} equals
View answer and explanation
Correct option: C
ii
Option C (ii) is correct because after substituting z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i\sin \theta ) and simplifying using the given condition ∣z−i∣=1|z - i| = 1, the expression cot⁡θ−2z\cot \mathrm{\theta } - \frac{2}{z} simplifies to ii.
672 Easy
672 of 979
Application +4 | -1
If z1{z_1} and z2{z_2} are two complex numbers such that ∣z1−z2z1+z2∣=1,\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} \right| = 1, then
View answer and explanation
Correct option: B
z1=i  k  z2,  k∈R{z_1} = i\;k\;{z_2},\;k \in R
Option B (z1=i;k;z2,;k∈R{z_1} = i\\;k\\;{z_2},\\;k \in R) is correct because the given condition implies that the diagonals of a parallelogram formed by z1z_1 and z2z_2 are equal, which means it's a rectangle, and thus z1z_1 and z2z_2 are orthogonal.
673 Medium
673 of 979
Application +4 | -1
The value of (1+i31−i3)6+(1−i31+i3)6\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right)^6+\left(\frac{1-i \sqrt{3}}{1+i \sqrt{3}}\right)^6 is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the expression simplifies to the sum of two complex numbers which are reciprocals of each other, resulting in 2.
674 Easy
674 of 979
Application +4 | -1
The complex number z=1+iz=1+i is rotated through an angle 3π/23 \pi / 2 in anti-clockwise direction about the origin and stretched by additional 2\sqrt{2} units, then the new complex number is
View answer and explanation
Correct option: D
None of the above
The correct answer is None of the above because after rotation and stretching, the new complex number is 2 - 2i.
675 Hard
675 of 979
Analyzing +4 | -1
If a=cos⁡(2π7)+isin⁡(2π7),  a = \cos \left( {\frac{{2\pi }}{7}} \right) + i\sin \left( {\frac{{2\pi }}{7}} \right),\; then the quadratic equation whose roots are α=a+a2+a4  and  β=a3+a5+a6\mathrm{\alpha } = a + {a^2} + {a^4}\;\mathrm{and}\;\mathrm{\beta } = {a^3} + {a^5} + {a^6}
View answer and explanation
Correct option: D
x2+x+2=0{x^2} + x + 2 = 0
Option D (x2+x+2=0{x^2} + x + 2 = 0) is correct because calculating the sum and product of the roots α and β leads to this quadratic equation.
676 Easy
676 of 979
Application +4 | -1
Which term of the sequence

(−8+18i),  (−6+15i),(−4+12i),  \left( { - 8 + 18i} \right),\;\left( { - 6 + 15i} \right),\left( { - 4 + 12i} \right),\; … is purely imaginary?
View answer and explanation
Correct option: A
5th
Option A (5th) is correct because the 5th term of the sequence is 0 + 6i, which is purely imaginary.
677 Easy
677 of 979
Application +4 | -1
If ∣z2−1∣=∣z∣2+1\left|z^2-1\right|=|z|^2+1, then P(z){P}(z) lies on
View answer and explanation
Correct option: C
The imaginary axis
Option C (The imaginary axis) is correct because by substituting z = x + iy into the given equation and simplifying, we find that x = 0, which means the point P(z) lies on the imaginary axis.
678 Medium
678 of 979
Application +4 | -1
Let α\alpha and β\beta be two fixed non-zero complex numbers and 'zz' a variable complex number. If the lines αzˉ+αˉz+1=0\alpha \bar z + \bar \alpha z + 1 = 0 and β  zˉ+βˉ  z−1=0\mathrm{\beta \;\bar z} + \bar \beta \;z - 1 = 0 are mutually perpendicular, then
View answer and explanation
Correct option: D
α  βˉ+αˉ  β=0\alpha \;\bar \beta + \bar \alpha \;\mathrm{\beta } = 0
Option D (α  βˉ+αˉ  β=0\alpha \;\bar \beta + \bar \alpha \;\mathrm{\beta } = 0) is correct because it represents the condition for the lines to be perpendicular, which involves the sum of the product of the coefficients and their conjugates being zero.
679 Medium
679 of 979
Application +4 | -1
Let p,q∈Rp, q \in \mathbb{R} and (1−3i)200=2199(p+iq),i=−1(1-\sqrt{3} i)^{200}=2^{199}(p+i q), i=\sqrt{-1}, then p+q+q2p+q+q^{2} and p−q+q2p-q+q^{2} are roots of the equation.
View answer and explanation
Correct option: A
x2−4x+1=0x^{2}-4 x+1=0
The correct answer is A (x2−4x+1=0x^{{2}} - 4x + 1 = 0) because it is the quadratic equation whose roots are derived from the given complex number equation.
680 Easy
680 of 979
Application +4 | -1
The smallest positive integer nn for which (1+i)2n=(1−i)2n{\left( {1 + i} \right)^{2n}} = {\left( {1 - i} \right)^{2n}} is
View answer and explanation
Correct option: C
2
Option C (2) is correct because it is the smallest positive integer n that satisfies the given equation, which simplifies to i2n=1{i^{2n}} = 1.
681 Easy
681 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}+i \frac{\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}
Option C (i3i \sqrt{{3}}) is correct because, after simplifying the powers of ω\omega and using the properties 1+ω+ω2=01 + \omega + \omega^2 = 0 and ω3=1\omega^3 = 1, the expression reduces to i3i \sqrt{{3}}.
682 Easy
682 of 979
Application +4 | -1
The values of xx and yy satisfying the equation (1+i)x−2i3+i+(2−3i)y+i3−i=i\frac{{\left( {1 + i} \right)x - 2i}}{{3 + i}} + \frac{{\left( {2 - 3i} \right)y + i}}{{3 - i}} = i are
View answer and explanation
Correct option: B
x=3,  y=−1x = 3,\;y = - 1
Option B (x=3,  y=−1x = 3,\;y = - 1) is correct because these values satisfy the given complex equation when substituted and simplified.
683 Easy
683 of 979
Application +4 | -1
The values of (16)1/4{\left( {16} \right)^{1/4}} are
View answer and explanation
Correct option: A
±2,  ±2i\pm 2,\; \pm 2i
The correct answer is A (±2,  ±2i\pm 2,\; \pm 2i) because the fourth roots of 16 are ±2\pm 2 and ±2i\pm 2i.
684 Easy
684 of 979
Analyzing +4 | -1
If the complex numbers z1,z2z_1, z_2 and z3z_3 represent the vertices of an equilateral triangle such that ∣z1∣=∣z2∣=∣z3∣\left|z_1\right|=\left|z_2\right|=\left|z_3\right|, then
View answer and explanation
Correct option: A
z1+z2+z3=0z_1+z_2+z_3=0
Option A (z1+z2+z3=0z_1+z_2+z_3=0) is correct because the centroid of an equilateral triangle with circumcenter at the origin is also at the origin.
685 Easy
685 of 979
Analyzing +4 | -1
If α,β,γ\alpha ,\beta ,\gamma are the cube roots of a negative number pp, then for any three real numbers x,  y,  zx,\;y,\;z the value of xα+yβ+zγxβ+yγ+zα\frac{{x\alpha + y\beta + z\gamma }}{{x\beta + y\gamma + z\alpha }} is
View answer and explanation
Correct option: B
−1−i32\frac{{ - 1 - i\sqrt 3 }}{2}
Option B (−1−i32\frac{{ - 1 - i\sqrt 3 }}{2}) is correct because simplifying the expression using the properties of cube roots of unity yields ω2\omega^2, which is equal to −1−i32\frac{{ - 1 - i\sqrt 3 }}{2}.
686 Easy
686 of 979
Understanding +4 | -1
The product of cube roots of −1- 1 is equal to
View answer and explanation
Correct option: A
−1- 1
Option A (−1- 1) is correct because the product of the cube roots of −1-1 is equal to −1-1.
687 Medium
687 of 979
Analyzing +4 | -1
The imaginary part of (z−1)(cos⁡α−isin⁡α)+(z−1)−1×(cos⁡α+isin⁡α)\left( {z - 1} \right)(\cos \alpha - i\sin \alpha ) + {\left( {z - 1} \right)^{ - 1}} \times (\cos \alpha + i\sin \alpha ) zero, if
View answer and explanation
Correct option: C
arg⁡(z−1)=α\arg \left( {z - 1} \right) = \mathrm{\alpha }
Option C (arg⁡(z−1)=α\arg \left( {z - 1} \right) = \mathrm{\alpha }) is correct because the imaginary part of the expression is zero when arg⁡(z−1)=α\arg(z-1) = \alpha.
688 Easy
688 of 979
Application +4 | -1
If ω\omega is a complex cube root of unity, then the equation ∣z−ω∣2+∣z−ω2∣2=λ{\left| {z - \omega } \right|^2} + {\left| {z - {\omega ^2}} \right|^2} = \lambda will represent a circle, if
View answer and explanation
Correct option: B
λ∈[3/2,∞)\mathrm{\lambda } \in \left[ {3/2,\infty } \right)
Option B (λ∈[3/2,∞)\mathrm{\lambda } \in \left[ {3/2,\infty } \right)) is correct because the condition for the equation to represent a circle is λ≥12∣z1−z2∣2\lambda \ge \frac{1}{2}{\left| {{z_1} - {z_2}} \right|^2}, where z1z_1 and z2z_2 are the centers.
689 Medium
689 of 979
Application +4 | -1
The value of ∑r=18(sin⁡2  r  π9+icos⁡2  r  π9),\sum \limits_{r = 1}^8 \left( {\sin \frac{{2\;r\;\pi }}{9} + i\cos \frac{{2\;r\;\pi }}{9}} \right), is
View answer and explanation
Correct option: D
−i- i
Option D (−i- i) is correct because it is the simplified value of the given summation after applying the formula for the sum of a geometric series and using properties of complex numbers.
690 Medium
690 of 979
Application +4 | -1
If z=−12+i32z=\frac{-1}{2}+i \frac{\sqrt{3}}{2}, then 8+10z+7z28+10 z+7 z^2 is equal to
View answer and explanation
Correct option: C
−12+i332\frac{-1}{2}+i \frac{3 \sqrt{3}}{2}
Option A ((0,2 tan-1 1/4]) is correct because it is the interval for 'a' where the discriminant of the quadratic equation is greater than zero.
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