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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Easy

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Practice format and chapter quiz

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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241 Easy
241 of 612
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.
242 Easy
242 of 612
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.
243 Easy
243 of 612
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.
244 Easy
244 of 612
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.
245 Easy
245 of 612
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.
246 Easy
246 of 612
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.
247 Easy
247 of 612
Application +4 | -1
The modulus and amplitude of (1+i3)8{\left( {1 + i\sqrt 3 } \right)^8} are respectively
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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}.
248 Easy
248 of 612
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.
249 Easy
249 of 612
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.
250 Easy
250 of 612
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.
251 Easy
251 of 612
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.
252 Easy
252 of 612
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}}.
253 Easy
253 of 612
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.
254 Easy
254 of 612
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.
255 Easy
255 of 612
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.
256 Easy
256 of 612
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 'b{b}' 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.
257 Easy
257 of 612
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.
258 Easy
258 of 612
Application +4 | -1
The curve represented by Re⁡(z2)=4\operatorname{Re}\left(z^2\right)=4 is
View answer and explanation
Correct option: D
a rectangular hyperbola
Option D (a rectangular hyperbola) is correct because the condition Re(z2z^2) = 4 translates to x2−y2=4x^2 - y^2 = 4, which is the equation of a rectangular hyperbola.
259 Easy
259 of 612
Application +4 | -1
The perpendicular distance of line (1−i)z+(1+i)zˉ+3=0(1-i) z+(1+i) \bar{z}+3=0 from (3+2i)(3+2 i) will be
View answer and explanation
Correct option: B
132\frac{13}{2}
Option B (132\frac{13}{2}) is correct because it's the perpendicular distance calculated using the formula for the distance of a point from a line in the complex plane.
260 Easy
260 of 612
Application +4 | -1
If z=1+iz=1+i, then the multiplicative inverse of z2z^2 is
View answer and explanation
Correct option: C
−i2-\frac{i}{2}
Option C (−i2-\frac{i}{2}) is correct because z2=(1+i)2=2iz^2 = (1+i)^2 = 2i, and the multiplicative inverse of 2i2i is 12i=−i2\frac{1}{2i} = -\frac{i}{2}.
261 Easy
261 of 612
Analyzing +4 | -1
The number of complex numbers satisfying zˉ=iz2\bar{z}=i z^2 is
View answer and explanation
Correct option: D
4
Option D (4) is correct because there are four complex numbers that satisfy the given equation: one where z=0 and three others derived from z3 = -i.
262 Easy
262 of 612
Application +4 | -1
If zz is a complex number such that Re (z)=Im  (z)\left( z \right) = \mathrm{Im}\;\left( z \right), then
View answer and explanation
Correct option: A
Re(z2)=0{\mathop\mathrm{Re}\nolimits} \left( {{z^2}} \right) = 0
Option A (Re(z2)=0{\mathop\mathrm{Re}\nolimits} \left( {{z^2}} \right) = 0) is correct because if the real and imaginary parts of z are equal, then the real part of z2 is zero.
263 Easy
263 of 612
Application +4 | -1
If α=cos⁡(8π11)+isin⁡(8π11)\alpha=\cos \left(\frac{8 \pi}{11}\right)+i \sin \left(\frac{8 \pi}{11}\right), then Re⁡(α+α2+α3+α4+α5)\operatorname{Re}\left(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5\right) is equal to
View answer and explanation
Correct option: B
−12-\frac{1}{2}
Option B (-12\frac{1}{2}) is correct because it is the real part of the sum of complex numbers on the unit circle.
264 Easy
264 of 612
Application +4 | -1
If 1,  ω,ω21,\;\omega ,{\omega ^2} are the cube roots of unity, then ω2(1+ω)3−(1+ω2)ω{\omega ^2}{\left( {1 + \omega } \right)^3} - \left( {1 + {\omega ^2}} \right)\omega is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because the expression simplifies to zero using the properties of cube roots of unity.
265 Easy
265 of 612
Analyzing +4 | -1
The value of the expression
(2−ω)(2−ω2)+2.(3−ω)(3−ω2)+...…+(n−1).(n−ω)(n−ω2)\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2.\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) + ... \ldots + \left( {n - 1} \right).\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)
Where ω\omega is an imaginary cube root of unity, is
View answer and explanation
Correct option: B
14(n−1)  n  (n2+3n+4)\frac{1}{4}\left( {n - 1} \right)\;n\;\left( {{n^2} + 3n + 4} \right)
Option B (14(n−1)n(n2+3n+4)\frac{1}{4}(n - 1)n(n^2 + 3n + 4)) is correct because it is the result of summing the series after simplifying the general term using properties of complex cube roots of unity.
266 Easy
266 of 612
Application +4 | -1
The area of the triangle whose vertices are represented by the complex number 0,z,zeiα,(0<α<π)0,z,z{e^{i\alpha }},(0 < \mathrm{\alpha } < \pi ) equals
View answer and explanation
Correct option: B
12∣z∣2sin⁡α\frac{1}{2}{\left| z \right|^2}\sin \mathrm{\alpha }
Option B (12∣z∣2sin⁡α\frac{1}{2}{\left| z \right|^2}\sin \mathrm{\alpha }) is correct because the area of the triangle can be found using the determinant formula with the given vertices, which simplifies to the expression involving the sine of the angle.
267 Easy
267 of 612
Analyzing +4 | -1
If zz satisfies the relation
∣z−i∣z1∣=∣z+i∣z1∣,|z-i| z 1|=| z+i|z 1|, (1)
then
View answer and explanation
Correct option: A
Im⁡(z)=0\operatorname{Im}(z)=0
Option A (Im⁡(z)=0\operatorname{Im}(z)=0) is correct because the relation implies that zz is equidistant from ii and −i-i when normalized, thus lying on the real axis.
268 Easy
268 of 612
Understanding +4 | -1
Which of the following is correct?
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Correct option: D
None of these
Option D (None of these) is correct because the order relation (>, <) is not defined for complex numbers.
269 Easy
269 of 612
Application +4 | -1
If Δ=∣6i−3i143i−1203i∣=x+iy\Delta=\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i\end{array}\right|=x+i y, then
View answer and explanation
Correct option: D
x=0,y=0x=0, y=0
Option D (x=0,y=0x=0, y=0) is correct because after applying the column operation, the determinant becomes 0.
270 Easy
270 of 612
Application +4 | -1
If ω(≠1)\omega(\neq 1) is a cube root of unity, then the value of tan⁡[(ω2017+ω2225)π−π/3]\tan \left[\left(\omega^{2017}+\omega^{2225}\right) \pi-\pi / 3\right]
View answer and explanation
Correct option: C
−3-\sqrt{3}
Option C (−3-\sqrt{3}) is correct because it is found by simplifying the expression inside the tangent function using properties of cube roots of unity.
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