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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Easy

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421 Easy
421 of 612
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}}.
422 Easy
422 of 612
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.
423 Easy
423 of 612
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.
424 Easy
424 of 612
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.
425 Easy
425 of 612
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}.
426 Easy
426 of 612
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.
427 Easy
427 of 612
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.
428 Easy
428 of 612
Application +4 | -1
The locus of zz satisying the inequality ∣z+2i2z+i∣<1\left| {\frac{{z + 2i}}{{2z + i}}} \right| < 1 where z=x+iyz = x + iy, is
View answer and explanation
Correct option: C
x2+y2>1{x^2} + {y^2} > 1
Option C (x2+y2>1{x^2} + {y^2} > 1) is correct because it represents the locus of points satisfying the given inequality, which simplifies to the exterior of a circle with radius 1 centered at the origin.
429 Easy
429 of 612
Application +4 | -1
The real part of (sin⁡x+icos⁡x)5(\sin x+\mathrm{i} \cos x)^5, where i=−1i=\sqrt{-1} is
View answer and explanation
Correct option: D
sin⁡5x\sin 5 x
Option D (sin⁡5x\sin 5 x) is correct because it represents the real part of the given complex expression.
430 Easy
430 of 612
Application +4 | -1
If f(z)=z4−8z3+4z2+4z+39f(z)=z^4-8 z^3+4 z^2+4 z+39 and f(3+2i)=a+ibf(3+2 i)=a+i b, then a:b=a: b=
View answer and explanation
Correct option: B
−1:8-1: 8
Option B (-1: 8) is correct because by dividing f(z)f(z) by z2−6z+13z^2-6z+13 and substituting z=3+2iz=3+2i, we find the ratio a:b is -1:8.
431 Easy
431 of 612
Understanding +4 | -1
Value of i4k+i4k+1+i4k+2+i4k+3i^{4 k}+i^{4 k+1}+i^{4 k+2}+i^{4 k+3} is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of consecutive powers of i with a difference of 1, starting from any power of 4, always equals zero.
432 Easy
432 of 612
Application +4 | -1
If α+iβ=tan⁡−1z,z=x+iy\alpha+i \beta=\tan ^{-1} z, z=x+i y and α\alpha is constant, then the locus of zz is
View answer and explanation
Correct option: A
x2+y2+2xcot⁡2α=1x^2+y^2+2 x \cot 2 \alpha=1
Option A (x2+y2+2xcot⁡2α=1x^2+y^2+2 x \cot 2 \alpha=1) is correct because it is derived by taking the tangent of both sides, separating real and imaginary parts, and manipulating the equation.
433 Easy
433 of 612
Application +4 | -1
The value of ∑k=110(sin⁡2kπ11+icos⁡2kπ11)\sum \limits_{k = 1}^{10} \left( {\sin \frac{{2k\pi }}{{11}} + i\cos \frac{{2k\pi }}{{11}}} \right) is
View answer and explanation
Correct option: C
−i- i
Option C (−i- i) is correct because the summation simplifies to −i-i using geometric series and Euler's formula.
434 Easy
434 of 612
Application +4 | -1
The value of S=∑k=16(sin⁡2πk7−icos⁡2πk7)S=\sum\limits_{{k=1}}^6\left(\sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7}\right)
View answer and explanation
Correct option: D
ii
Option D (i) is correct because by applying De Moivre's theorem and simplifying the summation, we obtain the result as i.
435 Easy
435 of 612
Application +4 | -1
∣12(z1+z2)+z1z2∣+∣12(z1+z2)−z1z2∣\left| {\frac{1}{2}\left( {{z_1} + {z_2}} \right) + \sqrt {{z_1}{z_2}} } \right| + \left| {\frac{1}{2}\left( {{z_1} + {z_2}} \right) - \sqrt {{z_1}{z_2}} } \right| is equal to
View answer and explanation
Correct option: C
∣z1∣+∣z2∣\left| {{z_1}} \right| + \left| {{z_2}} \right|
Option C (∣z1∣+∣z2∣\left| {{{{z_1}}}} \right| + \left| {{{{z_2}}}} \right|) is correct because it is the simplified form of the given expression using properties of complex numbers and their magnitudes.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs DIFFERENTIAL EQUATIONS 977 MCQs
436 Easy
436 of 612
Understanding +4 | -1
Statement-I: If (a+ib)(c+id)(e+if)(g+ih)=A+iB(a+i b)(c+i d)(e+if )(g+i h)=A+i B, then (a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=A^2+B^2.

Statement-II: If z=x+iy,z=x+ iy, then ∣z∣=x2+y2|z|=\sqrt{x^2+y^2}.
View answer and explanation
Correct option: C
Both are correct
Option C is correct because both statements are correct. Statement I relates complex number multiplication to the product of their magnitudes, and Statement II defines the modulus of a complex number.
437 Easy
437 of 612
Application +4 | -1
The solution of log⁡99{log⁡2(log⁡3x)}=0{\log _{99}}\left\{ {{{\log }_2}\left( {{{\log }_3}x} \right)} \right\} = 0 is
View answer and explanation
Correct option: B
9
Option B (9) is correct because it is the solution to the given logarithmic equation after simplifying step by step.
438 Easy
438 of 612
Application +4 | -1
If (1+i1−i)n=−1,n∈N\left(\frac{1+i}{1-i}\right)^n=-1, \mathrm{n} \in \mathbf{N}, then least value of nn is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the least positive integer value of n that satisfies the equation is 2, as (1+i)/(1-i) simplifies to i, and i2 = -1.
439 Easy
439 of 612
Application +4 | -1
The value of ∑n=0∞(2i3)n\sum \limits_{n = 0}^\infty {\left( {\frac{{2i}}{3}} \right)^n} is
View answer and explanation
Correct option: A
9+6i13\frac{{9 + 6i}}{{13}}
Option A ((9 + 6i)/13) is correct because the given expression is an infinite geometric series, and its sum can be calculated using the formula for the sum of an infinite geometric series.
440 Easy
440 of 612
Application +4 | -1
The complex numbers z1,z2,z3z_1, z_2, z_3 are the vertices A,B,C{A}, {B}, {C} of a parallelogram ABCDA B C D, then the fourth vertex DD is
View answer and explanation
Correct option: D
z1+z3−z2z_1+z_3-z_2
Option D (z1+z3−z2z_1+z_3-z_2) is correct because the fourth vertex of a parallelogram can be found using the property that diagonals bisect each other.
441 Easy
441 of 612
Application +4 | -1
If ω\omega is an imaginary cube root of unity, then (1+ω−ω2)7{\left( {1 + \omega - {\omega ^2}} \right)^7} equals
View answer and explanation
Correct option: D
−128  ω2- 128\;{\omega ^2}
Option D (−128  ω2- 128\;{\omega ^2}) is correct because simplifying the expression using the properties of cube roots of unity leads to this result.
442 Easy
442 of 612
Application +4 | -1
If x+iy=11−cos⁡θ+2isin⁡θ,θ≠2nπ,n∈Ix+i y=\frac{1}{1-\cos \theta+2 i \sin \theta}, \theta \neq 2 n \pi, n \in \boldsymbol{I}, then maximum value of xx is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{{1}}{{2}}) is correct because the maximum value of x occurs when cos⁡θ=−1\cos \theta = -1, resulting in x=12x = \frac{{1}}{{2}}.
443 Easy
443 of 612
Analyzing +4 | -1
Let z1,z2,z3{z_1},{z_2},{z_3} be three complex numbers satisfying 1z1+1z2+1z3=0.\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + \frac{1}{{{z_3}}} = 0. Let zk=rk(cos⁡αk+isin⁡αk){z_k} = {r_k}\left( {\cos {\alpha _k} + i\sin {\alpha _k}} \right) and ωk=cos⁡2αk+isin⁡2  αkzk{\omega _k} = \frac{{\cos 2{\alpha _k} + i\sin 2\;{\alpha _k}}}{{{z_k}}} for k=1,2,3.k = 1,2,3. If ω1,ω2{\omega _1},{\omega _2} and ω3{\omega _3} are the affixes of points A1,A2{A_1},{A_2} and A3{A_3} respectively in the Argand plane, then ΔA1A2A3\Delta {A_1}{A_2}{A_3} has its
View answer and explanation
Correct option: B
Centroid at the origin
Option B (Centroid at the origin) is correct because the sum of the complex numbers representing the vertices is zero, implying the centroid is at the origin.
444 Easy
444 of 612
Application +4 | -1
If Re⁡(z+2iz+4)=0\operatorname{Re}\left(\frac{z+2 i}{z+4}\right)=0, then zz lies on a circle with centre
View answer and explanation
Correct option: A
(−2,−1)(-2,-1)
Option A ((-2,-1)) is correct because the real part of the given expression being zero implies that z lies on a circle with center (-2, -1).
445 Easy
445 of 612
Analyzing +4 | -1
If ω\omega is the complex cube root of unity, then the value of ω+ω(12+38+932+27128+…)\omega + {\omega ^{\left( {\frac{1}{2} + \frac{3}{8} + \frac{9}{{32}} + \frac{{27}}{{128}} + \ldots } \right)}}
View answer and explanation
Correct option: A
−1- 1
Option A (-1) is correct because the infinite geometric series simplifies to 2, and ω + ω² = -1.
446 Easy
446 of 612
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: B
2
Option B (2) is correct because it is the smallest positive integer that satisfies the given equation after simplification using properties of complex numbers.
447 Easy
447 of 612
Analyzing +4 | -1
The solution set of the inequation ∣x+1x∣>2,\left| {x + \frac{1}{x}} \right| > 2, is
View answer and explanation
Correct option: B
R−{−1,0,1}R - \left\{ { - 1,0,1} \right\}
Option B (R−−1,0,1R - \\{ - 1,0,1 \\}) is correct because it excludes the values where the expression inside the absolute value is equal to 2 or -2, and also excludes 0 where the expression is undefined.
448 Easy
448 of 612
Analyzing +4 | -1
If the equation, x2+bx+45=0  (b ∈R){x^2} + bx + 45 = 0\,\,\left( {b\, \in R} \right) has conjugate complex roots and they satisfy ∣z+1∣=210,\left| {z + 1} \right| = 2\sqrt {10} , then
View answer and explanation
Correct option: C
b2−b=30{b^2} - b = 30
Option C (b2−b=30{b^2} - b = 30) is correct because by using the given condition on the complex roots, we can find the value of bb and then calculate b2−b{b^2} - b.
449 Easy
449 of 612
Analyzing +4 | -1
The complex numbers sin⁡x+icos⁡2x\sin x + i\cos 2x and cos⁡x−isin⁡2x\cos x - i\sin 2x are conjugate to each other for
View answer and explanation
Correct option: D
No value of xx
Option D (No value of x) is correct because there is no value of x that satisfies both sin(x) = cos(x) and cos(2x) = -sin(2x) simultaneously.
450 Easy
450 of 612
Application +4 | -1
If α\alpha and β\beta are the complex cube roots of unity, then α4+β4+α−1β−1=\alpha^4+\beta^4+\alpha^{-1} \beta^{-1}=
View answer and explanation
Correct option: A
0
Option A (0) is correct because, after substituting α=ω\alpha = \omega and β=ω2\beta = \omega^2 and simplifying using the properties of cube roots of unity, the expression evaluates to 0.
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