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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Mixed difficulty

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691 Easy
691 of 979
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.
692 Easy
692 of 979
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.
693 Easy
693 of 979
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.
694 Easy
694 of 979
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.
695 Medium
695 of 979
Application +4 | -1
If z=x+iyz = x + iy, then area of the triangle whose vertices are points z,  iz,  z+izz,\;iz,\;z + iz is
View answer and explanation
Correct option: A
12∣z∣2\frac{1}{2}{\left| z \right|^2}
Option A (12∣z∣2\frac{1}{2}{\left| z \right|^2}) is correct because the area of the triangle formed by the given vertices is half the square of the magnitude of z.
696 Medium
696 of 979
Application +4 | -1
If (1+cos⁡ϕ+isin⁡ϕ1+cos⁡ϕ−isin⁡ϕ)=u+iv\left( {\frac{{1 + \cos \phi + i\sin \phi }}{{1 + \cos \phi - i\sin \phi }}} \right) = u + iv, where uu and vv all real numbers, then uu is
View answer and explanation
Correct option: B
cos⁡nϕ\cos n\phi
Option B (cos⁡nϕ\cos n\phi) is correct because the expression simplifies to cos⁡nϕ+isin⁡nϕ\cos n\phi + i\sin n\phi, and thus u = cos⁡nϕ\cos n\phi.
697 Easy
697 of 979
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.
698 Easy
698 of 979
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.
699 Easy
699 of 979
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.
700 Easy
700 of 979
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.
701 Easy
701 of 979
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.
702 Easy
702 of 979
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.
703 Easy
703 of 979
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.
704 Easy
704 of 979
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.
705 Medium
705 of 979
Application +4 | -1
The centre of a regular hexagon is at the point z=iz = i. If one of its vertices is at 2+i2 + i, then the adjacent vertices of 2+i2 + i are at the points
View answer and explanation
Correct option: D
1+i(1±3)1 + i\left( {1 \pm \sqrt 3 } \right)
Option D (1+i(1±3)1 + i\left( {1 \pm \sqrt 3 } \right)) is correct because it represents the vertices adjacent to 2+i2+i in a regular hexagon centered at ii.

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706 Easy
706 of 979
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.
707 Medium
707 of 979
Application +4 | -1
If z=x−iyz=x-i y and z13=p+iqz^{\frac{1}{3}}=p+i q, then (xp+yq)/(p2+q2)\left(\frac{x}{p}+\frac{y}{q}\right) /\left(p^{2}+q^{2}\right) is equal to
View answer and explanation
Correct option: A
-2
Option A (-2) is correct because by cubing z13=p+iqz^{\frac{1}{3}} = p + iq and equating real and imaginary parts, the expression simplifies to -2.
708 Easy
708 of 979
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.
709 Easy
709 of 979
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}}.
710 Easy
710 of 979
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.
711 Medium
711 of 979
Application +4 | -1
If z=x+iyz = x + iy is a variable complex number such that arg⁡z−1z+1=π4\arg \frac{{z - 1}}{{z + 1}} = \frac{\pi }{4}, then
View answer and explanation
Correct option: C
x2+y2−2y=1{x^2} + {y^2} - 2y = 1
Option C (x2+y2−2y=1{x^2} + {y^2} - 2y = 1) is correct because by substituting z=x+iyz = x + iy into the given argument equation and simplifying, we obtain the equation of a circle.
712 Medium
712 of 979
Application +4 | -1
If z1,z2\mathrm{z}_{1}, \mathrm{z}_{2} and z3,z4\mathrm{z}_{3}, \mathrm{z}_{4} are 2 pairs of complex conjugate numbers, then
arg⁡(z1z4)+arg⁡(z2z3)\arg \left(\frac{z_{1}}{z_{4}}\right)+\arg \left(\frac{z_{2}}{z_{3}}\right) equals:
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the arguments of a complex number and its conjugate is zero, and z2 and z4 are conjugates of z1 and z3 respectively.
713 Medium
713 of 979
Application +4 | -1
If n∈Nn \in \mathbf{N}, then (1+i)n(1−i)n−2\frac{(1+i)^n}{(1-i)^{n-2}} is equal to
View answer and explanation
Correct option: B
−2in+1-2 i^{n+1}
Option B (−2in+1-2 i^{n+1}) is correct because after simplifying the expression using properties of complex numbers, the result matches this option.
714 Easy
714 of 979
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).
715 Easy
715 of 979
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.
716 Easy
716 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: 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.
717 Easy
717 of 979
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.
718 Easy
718 of 979
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.
719 Easy
719 of 979
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.
720 Medium
720 of 979
Application +4 | -1
If Z{Z} is a unimodular complex number then arg⁡(z2+zˉ)=\arg \left(z^2+\bar{z}\right)=
View answer and explanation
Correct option: A
12arg⁡z\frac{1}{2} \arg z
Option A (12arg⁡z\frac{1}{2} \arg z) is correct because after simplification using trigonometric identities, the argument of the expression reduces to half of the argument of z.
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