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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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391 Medium
391 of 979
Application +4 | -1
If (1+z)n=a0+a1z+a2z2+⋯+anzn(1+z)^n=a_0+a_1 z+a_2 z^2+\cdots+a_n z^n, where a0,a1a_0, a_1, a2,…,ana_2, \ldots, a_n, are real, then (a0−a2+a4−a6+⋯ )2+(a1−a3+a5−a7+⋯ )2=\left(a_0-a_2+a_4-a_6+\cdots\right)^2+\left(a_1-a_3+a_5-a_7+\cdots\right)^2=
View answer and explanation
Correct option: A
2n2^n
Option A (2n2^n) is correct because substituting z=i into the binomial expansion and comparing magnitudes leads to this result.
392 Easy
392 of 979
Application +4 | -1
If log⁡107=0.8451{\log _{10}}7 = 0.8451, then the position of the first significant figure of 7−20{7^{ - 20}} is
View answer and explanation
Correct option: B
17
Option B (17) is correct because the position of the first significant figure is determined by the characteristic of the logarithm of the number.
393 Easy
393 of 979
Understanding +4 | -1
Reflection of the line aˉz+azˉ=0\bar{a} z+a \bar{z}=0 in the real axis is
View answer and explanation
Correct option: A
aˉzˉ+az=0\bar{a} \bar{z}+a z=0
Option A (aˉzˉ+az=0\bar{a} \bar{z}+a z=0) is correct because reflecting a complex number in the real axis means taking its conjugate, which transforms the equation accordingly.
394 Easy
394 of 979
Application +4 | -1
The complex number i+3i+\sqrt{3} in polar form can be written as
View answer and explanation
Correct option: B
2(cos⁡π6+isin⁡π6)2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)
Option B is correct because the complex number i+3i+\sqrt{3} can be expressed in polar form as 2(cos⁡π6+isin⁡π6)2(\cos{\frac{\pi}{6}}+i \sin{\frac{\pi}{6}}).
395 Hard
395 of 979
Application +4 | -1
The maximum distance from the origin of coordinates to the point zz satisfying the equation ∣z+1z∣=a\left| {z + \frac{1}{z}} \right| = a is
View answer and explanation
Correct option: C
12(a2+4+a)\frac{1}{2}\left( {\sqrt {{a^2} + 4} + a} \right)
Option C 12(a2+4+a)\frac{{1}}{{2}}\left( {\sqrt {{{{a^2} + 4}}} + a} \right) is correct because it represents the maximum distance derived by maximizing the modulus of the complex number and solving the resulting quadratic equation.
396 Easy
396 of 979
Application +4 | -1
If the real part of zˉ+2zˉ−1\frac{{\bar z + 2}}{{\bar z - 1}} is 4, then the locus of the point representing zz in the complex plane is
View answer and explanation
Correct option: A
a circle
Option A (a circle) is correct because the locus of the point representing z is a circle, derived from the given condition on the real part of the complex expression.
397 Easy
397 of 979
Application +4 | -1
Let z1,z2z_1, z_2 be two complex numbers such that z1≠0z_1 \neq 0 and z2/z1z_2 / z_1 is purely real, then ∣2iz1+5z22iz1−5z2∣\left|\frac{2 i z_1+5 z_2}{2 i z_1-5 z_2}\right| is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because if z_2/z_1 is purely real, the magnitude of the given expression simplifies to 1.
398 Medium
398 of 979
Application +4 | -1
If z=x+iy,xy≠0z=x+i y, x y \neq 0, satisfies the equation z2+izˉ=0z^2+i \bar{z}=0, then ∣z2∣\left|z^2\right| is equal to :
View answer and explanation
Correct option: C
1
The correct answer is C (1) because by taking the modulus of the given equation, we find that |z| = 1, and therefore |z2| = 1.
399 Easy
399 of 979
Understanding +4 | -1
A point PP which represents a complex number zz moves such that ∣z−z1∣=∣z−z2∣\left| {z - {z_1}} \right| = \left| {z - {z_2}} \right|, then its locus is
View answer and explanation
Correct option: D
Perpendicular bisector of line joining z1{z_1} and z2{z_2}
Option A (zzˉ−zzˉ0−zˉz0+z0zˉ0=r2z\bar z - z{\bar z_0} - \bar z{z_0} + {z_0}{\bar z_0} = {r^2}) is correct because it represents the equation of a circle in the complex plane with center z0 and radius r.
400 Easy
400 of 979
Analyzing +4 | -1
For n∈Nn \in N, let Sn={z∈C:∣z−3+2i∣=n4}S_n=\left\{z \in C:|z-3+2 i|=\frac{n}{4}\right\} and Tn={z∈C:∣z−2+3i∣=1n}T_n=\left\{z \in C:|z-2+3 i|=\frac{1}{n}\right\}. Then the number of elements in the set {n∈N:Sn∩Tn=ϕ}\left\{n \in N: S_n \cap T_n=\phi\right\} is
View answer and explanation
Correct option: B
∞\infty
Option B (∞\infty) is correct because the intersection of the two circles is empty for an infinite number of values of n.
401 Easy
401 of 979
Application +4 | -1
The modulus of 2i−−2i\sqrt{2{i}}-\sqrt{-2 {i}} is:
View answer and explanation
Correct option: A
2
Option A (2) is correct because simplifying the expression 2i−−2i\sqrt{2i} - \sqrt{-2i} and then finding its modulus results in 2.
402 Easy
402 of 979
Application +4 | -1
The value of (cos⁡2θ−isin⁡2θ)7(cos⁡3θ+isin⁡3θ)−5(cos⁡4θ+isin⁡4θ)12(cos⁡5θ+isin⁡5θ)−6\frac{(\cos 2 \theta-i \sin 2 \theta)^7(\cos 3 \theta+i \sin 3 \theta)^{-5}}{(\cos 4 \theta+i \sin 4 \theta)^{12}(\cos 5 \theta+i \sin 5 \theta)^{-6}} is
View answer and explanation
Correct option: D
cos⁡47θ−isin⁡47θ\cos 47 \theta-i \sin 47 \theta
Option D (cos⁡47θ−isin⁡47θ\cos 47 \theta - i \sin 47 \theta) is correct because after applying De Moivre's theorem and simplifying, the expression equals cos⁡47θ−isin⁡47θ\cos 47 \theta - i \sin 47 \theta.
403 Easy
403 of 979
Application +4 | -1
Let Z=[7+5i3+3i8+5i5+2i]Z=\left[\begin{array}{ll}7+5 i & 3+3 i \\ 8+5 i & 5+2 i\end{array}\right] then det⁡(ZZˉ)\operatorname{det}(Z \bar{Z}), (where Zˉ\bar{Z} is the matrix obtained by taking conjugate of each element of ZZ ) is equal to
View answer and explanation
Correct option: A
256
Option A (256) is correct because the determinant of Z * conjugate(Z) is the square of the magnitude of the determinant of Z.
404 Easy
404 of 979
Application +4 | -1
If x+iy=32+cos⁡θ+isin⁡θ,   then   x2+y2x + iy = \frac{3}{{2 + \cos \mathrm{\theta } + i\sin \mathrm{\theta }}},\;\,\mathrm{then}\,\;{x^2} + {y^2} is equal to
View answer and explanation
Correct option: B
4  x−34\;x - 3
Option B (4x−34x - 3) is correct because after rationalizing the denominator and equating real and imaginary parts, x2+y2{x^2} + {y^2} simplifies to 4x−34x - 3.
405 Medium
405 of 979
Application +4 | -1
If x1≤x2x_1 \leq x_2 and y1≤y2y_1 \leq y_2 for any two complex numbers z1=x1+iy1z_1=x_1+i y_1 and z2=x2+iy2z_2=x_2+i y_2, then we say z1∩z2.z_1 \cap z_2.

Let zz be a complex number with 1∩z1 \cap z. Then
View answer and explanation
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 the condition 1 ∩ z, it follows that the real and imaginary parts of 1−z1+z\frac{{1-z}}{{1+z}} are non-positive, implying it is \"less than or equal to\" 0.
406 Easy
406 of 979
Application +4 | -1
If x+iy=32+cos⁡θ+isin⁡θx+i y=\frac{3}{2+\cos \theta+i \sin \theta} then x2+y2=x^2+y^2=
View answer and explanation
Correct option: A
4x−34 x-3
Option A (4x−34x-3) is correct because by expressing x and y in terms of θ and then calculating x2+y2x^2 + y^2, it can be shown that x2+y2=4x−3x^2 + y^2 = 4x - 3.
407 Medium
407 of 979
Analyzing +4 | -1
If p,q,rp,q,r are positive and are in AP, then roots of the quadratic equation px2+qx+r=0p{x^2} + qx + r = 0 are complex for
View answer and explanation
Correct option: B
∣pr−7∣<43\left| {\frac{p}{r} - 7} \right| < 4\sqrt 3
Option B is correct because the condition for complex roots of the quadratic equation, given that p, q, and r are in AP, leads to the inequality ∣pr−7∣<43\left| {\frac{{p}}{{r}} - 7} \right| < 4\sqrt 3.
408 Easy
408 of 979
Application +4 | -1
i2+i4+i6+...upto  (2k+1)terms,  k∈N{i^2} + {i^4} + {i^6} + ...\mathrm{upto}\;\left( {2k + 1} \right)\mathrm{terms},\;k \in N is
View answer and explanation
Correct option: C
−1- 1
Option C (-1) is correct because the series simplifies to a geometric series with a common ratio of -1, and the sum is -1.
409 Easy
409 of 979
Application +4 | -1
The modulus and amplitude of (1+i3)8{\left( {1 + i\sqrt 3 } \right)^8} are respectively
View answer and explanation
Correct option: B
256 and 2π3\frac{{2\pi }}{3}
Option B (256 and 2π3\frac{{2\pi }}{3}) is correct because (1+i3)8(1 + i\sqrt 3)^8 can be simplified using De Moivre's theorem, resulting in a modulus of 256 and an argument of 2π3\frac{{2\pi }}{3}.
410 Easy
410 of 979
Application +4 | -1
If x=log⁡ba,  y=log⁡cb,  z=log⁡acx = {\log _b}a,\;y = {\log _c}b,\;z = {\log _a}c, then xyzxyz is
View answer and explanation
Correct option: B
1
Option B (1) is correct because when you multiply the logarithms together, the terms cancel out, leaving a result of 1.
411 Easy
411 of 979
Analyzing +4 | -1
Let z1   and   z2{z_1}{\text{ }}\,\,{\text{and}}\,\,{\text{ }}{z_2} be two complex numbers such that z1‾=i z2‾   and   arg⁡(z1z2‾)=π.\overline {{z_1}} = i\,\overline {{z_2}} {\text{ }}\,\,{\text{and }}\,\,\arg \left( {\frac{{{z_1}}}{{\overline {{z_2}} }}} \right) = \pi . then
View answer and explanation
Correct option: C
arg⁡ z1=π4\arg \,{z_1} = \frac{\pi }{4}
Option C (arg z₁ = π/4) is correct because by using the given relationships between z₁ and z₂ and their arguments, we can solve for arg z₁.
412 Easy
412 of 979
Understanding +4 | -1
Consider the following statements.

I. Additive inverse of (1−i)(1-i) is equal to −1+i-1+i.

II. If z1{z}_1 and z2{z}_2 are two complex numbers, then z1−z2{z}_1-{z}_2 represents a complex number which is sum of z1z_1 and additive inverse of z2z_2.
III. Simplest form of 5+2i1−2i\frac{5+\sqrt{2} i}{1-\sqrt{2} i} is 1+22i.1+2 \sqrt{2}i.

Choose the correct option.
View answer and explanation
Correct option: C
I, II and III are correct
Option C is correct because all three statements are correct. Statement I correctly identifies the additive inverse, Statement II describes complex number subtraction, and Statement III simplifies the given expression correctly.
413 Easy
413 of 979
Understanding +4 | -1
The inequality a+ib>c+ida+i b > c+i d is true when
View answer and explanation
Correct option: B
a>c,b=d=0a > c, b=d=0
Option B (a>c,b=d=0a > c, b=d=0) is correct because inequalities are only defined for real numbers, thus the imaginary parts must be zero, and the real parts must satisfy the inequality.
414 Easy
414 of 979
Application +4 | -1
If ω\omega is a complex cube root of unity, then sin⁡{(ω10+ω23π−π4)}\sin \left\{ {\left( {{\omega ^{10}} + {\omega ^{23}}\pi - \frac{\pi }{4}} \right)} \right\} is equal to
View answer and explanation
Correct option: A
12\frac{1}{{\sqrt 2 }}
Option A (12\frac{1}{\sqrt{2}}) is correct because simplifying the expression inside the sine function using properties of complex cube roots of unity leads to sin⁡(−π−π4)=12\sin(-\pi - \frac{\pi}{4}) = \frac{1}{\sqrt{2}}.
415 Easy
415 of 979
Application +4 | -1
The locus of point zz satisfying Re(1z)=k{\mathop\mathrm{Re}\nolimits} \left( {\frac{1}{z}} \right) = k, where kk is a non-zero real number, is
View answer and explanation
Correct option: B
A circle
Option B (A circle) is correct because the equation Re(1z)=k{\mathop\mathrm{Re}\nolimits} \left( {\frac{{1}}{{z}}} \right) = k simplifies to the equation of a circle.
416 Easy
416 of 979
Application +4 | -1
If 32+cos⁡θ+isin⁡θ=a+ib\frac{3}{{2 + \cos \theta + i\sin \theta }} = a + ib, then [(a−2)2+b2]\left[ {{{\left( {a - 2} \right)}^2} + {b^2}} \right] is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because after rationalizing the denominator and separating real and imaginary parts, the expression (a−2)2+b2{{\left( {a - 2} \right)}^2} + {b^2} simplifies to 1.
417 Easy
417 of 979
Application +4 | -1
If 1,  ω,ω21,\;\omega ,{\omega ^2} are the cube roots of unity, then Δ=∣1ωnω2nωnω2n1ω2n1ωn∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} 1&{{\omega ^n}}&{{\omega ^{2n}}}\\ {{\omega ^n}}&{{\omega ^{2n}}}&1\\ {{\omega ^{2n}}}&1&{{\omega ^n}} \end{array}} \right| is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the determinant simplifies to 0 using the properties of cube roots of unity, where ω3n=1\omega^{3n} = 1.
418 Easy
418 of 979
Application +4 | -1
If z=5i(−35i)z=5 i\left(\frac{-3}{5} i\right), then zz is equal to 3+bi.3+bi. The value of &#39;b{b}&#39; is
View answer and explanation
Correct option: A
0
Option A (0) is correct because simplifying the expression results in 3 + 0i, so the value of 'b' is 0.
419 Medium
419 of 979
Application +4 | -1
e2micot⁡−1p⋅(pi+1pi−1)me^{2 m i \cot ^{-1} p} \cdot\left(\frac{p i+1}{p i-1}\right)^m is equal to
View answer and explanation
Correct option: B
1
The expression e2micot⁡−1p⋅(pi+1pi−1)me^{2 m i \cot ^{-1} p} \cdot\left(\frac{{p i+1}}{{p i-1}}\right)^m is equal to 1 because it simplifies to 1 using trigonometric identities and Euler's formula.
420 Easy
420 of 979
Application +4 | -1
If z=3−i2z=\frac{\sqrt{ } 3-i}{2}, where i=−1i=\sqrt{-1}, then (i101+z101)103\left(i^{101}+z^{101}\right)^{103} equals to
View answer and explanation
Correct option: B
zz
Option B (z) is correct because by substituting the given value of z and simplifying using properties of complex numbers and cube roots of unity, the expression simplifies to z.
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