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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Mixed difficulty

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91 Medium
91 of 979
Analyzing +4 | -1
If z1,  z2,  …,zn{z_1},\;{z_2},\; \ldots ,{z_n} are complex numbers such that ∣z1∣=∣z2∣=...=∣zn∣=1\left| {{z_1}} \right| = \left| {{z_2}} \right| = ... = \left| {{z_n}} \right| = 1, then ∣z1+z2+...+zn∣\left| {{z_1} + {z_2} + ... + {z_n}} \right| is equal to
View answer and explanation
Correct option: C
∣1z1+1z2+...+1zn∣\left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ... + \frac{1}{{{z_n}}}} \right|
Option C is correct. Given the modulus of each complex number is 1, the expression ∣z1+z2+...+zn∣\left| {{z_1} + {z_2} + ... + {z_n}} \right| is equal to ∣1z1+1z2+...+1zn∣\left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ... + \frac{1}{{{z_n}}}} \right| because the conjugate of each ziz_i is its reciprocal.
92 Easy
92 of 979
Application +4 | -1
Find Re⁡(z)\operatorname{Re}(z), which satisfy ∣z∣=1|z|=1 and z−2z=1z-\frac{2}{z}=1
View answer and explanation
Correct option: B
−1-1
Option B (-1) is correct. By substituting z = x + yi and using the given conditions, we find that Re(z) = x = -1.
93 Easy
93 of 979
Analyzing +4 | -1
If z1,z2,z3{z_1},{z_2},{z_3} be vertices of an equilateral triangle occurring in the anticlockwise sense, then
View answer and explanation
Correct option: C
z1+ω  z2+ω2z3=0{z_1} + \omega \;{z_2} + {\omega ^2}{z_3} = 0
Option D (−2,2,4- 2,\mathrm{ }2,\mathrm{ }4) is correct because these are the solutions obtained by considering both positive and negative cases of the absolute value and verifying that they satisfy the original equation.
94 Easy
94 of 979
Application +4 | -1
For n=6  k,  k∈Z,(1−i32)n+(−1−i−32)nn = 6\;k,\;k \in Z,{\left( {\frac{{1 - i\sqrt 3 }}{2}} \right)^n} + {\left( {\frac{{ - 1 - i - \sqrt 3 }}{2}} \right)^n} has the value
View answer and explanation
Correct option: D
2
The correct answer is D (2) because substituting n=6kn = 6k into the expression and using the properties of complex cube roots of unity simplifies the expression to 2.
95 Easy
95 of 979
Application +4 | -1
The value of cos⁡30∘+isin⁡300cos⁡60∘−isin⁡600\frac{{\cos 30^\circ + i\sin {{30}^0}}}{{\cos 60^\circ - i\sin {{60}^0}}} is equal to
View answer and explanation
Correct option: A
ii
Option A (i) is correct because using Euler's formula and simplifying the expression results in the imaginary unit i.
96 Medium
96 of 979
Application +4 | -1
The root of the equation 2(1+i)x2−4(2−i)x−5−3i=0,2\left( {1 + i} \right){x^2} - 4\left( {2 - i} \right)x - 5 - 3i = 0, where i=−1,i = \sqrt { - 1} , which has greater modulus, is
View answer and explanation
Correct option: A
3−5i2\frac{{3 - 5i}}{2}
Option A (3−5i2\frac{{3 - 5i}}{2}) is correct because it is one of the roots of the quadratic equation and has a greater modulus than the other root.
97 Medium
97 of 979
Analyzing +4 | -1
The number of complex numbers zz such that ∣z∣<1/3|z| < 1 / 3, and ∑r=1narzr=1\sum\limits_{{r=1}}^n a_r z^r=1 where ∣ar∣<2\left|a_r\right| < 2, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the given condition leads to a contradiction, implying there are no such complex numbers z.
98 Easy
98 of 979
Analyzing +4 | -1
All the points in the set S={α+iα−i:α∈R}(i=−1)S = \left\{ {\frac{{\alpha + i}}{{\alpha - i}}:\alpha \in R} \right\}(i = \sqrt { - 1} ) lie on a
View answer and explanation
Correct option: C
circle whose radius is 1.
Option C (circle whose radius is 1) is correct because the given expression represents a complex number whose magnitude is always 1, thus lying on a circle of radius 1 centered at the origin.
99 Easy
99 of 979
Analyzing +4 | -1
If aa and bb are real numbers between 0 and 1 such that the points z1=a+i,z2=1+biz_1=a+i, z_2=1+b i and z3=0z_3=0 form an equilateral triangle, then
View answer and explanation
Correct option: A
a=b=2−3a=b=2-\sqrt{3} should look like this
Option A (a=b=2−3a=b=2-\sqrt{3}) is correct because it satisfies the equilateral triangle condition and the constraint that a and b are between 0 and 1.
100 Easy
100 of 979
Application +4 | -1
If z1,  z2{z_1},\;{z_2} and z3{z_3} are any three complex numbers, then the fourth vertex of the parallelogram whose three vertices are z1,  z2{z_1},\;{z_2} and z3{z_3} taken in order is
View answer and explanation
Correct option: A
z1−z2+z3{z_1} - {z_2} + {z_3}
Option A (z1−z2+z3{z_1} - {z_2} + {z_3}) is correct because in a parallelogram, the midpoints of the diagonals coincide, allowing us to find the fourth vertex.
101 Easy
101 of 979
Application +4 | -1
If ∣z−3−2i∣=∣z+2i∣|z-3-2 i|=|z+2 i|, where zz is a complex number, then minimum value of ∣z∣|z| will be
View answer and explanation
Correct option: D
910\frac{9}{10}
Option D (910\frac{9}{10}) is correct because the equation represents a straight line, and the minimum distance from the origin to the line is 910\frac{9}{10}.
102 Easy
102 of 979
Application +4 | -1
kk non-overlapping regular polygons of p1,p2,…,pkp_1, p_2, \ldots, p_k sides have one common vertex so that no gap is left at that vertex. Then 1p1+1p2+⋯+1pk\frac{1}{p_1}+\frac{1}{p_2}+\cdots+\frac{1}{p_k} is equal to
View answer and explanation
Correct option: B
k2−1\frac{k}{2}-1
The value of (1+isin⁡π8+cos⁡π81−isin⁡π8+cos⁡π8)8\left(\frac{{1+i \sin \frac{{\pi}}{{8}}+\cos \frac{{\pi}}{{8}}}}{{1-i \sin \frac{{\pi}}{{8}}+\cos \frac{{\pi}}{{8}}}}\right)^8 is -1 because it simplifies to -1 using trigonometric identities and De Moivre's theorem.
103 Medium
103 of 979
Analyzing +4 | -1
Let α=ei5π21\alpha=e^{i \frac{5 \pi}{21}} and z=α−α2+α3−α4+…..+α9−α10z=\alpha-\alpha^2+\alpha^3-\alpha^4+\ldots . .+\alpha^9-\alpha^{10}. If A=Re⁡(z)A=\operatorname{Re}({z}), then ∣A∣|{A}| is equal to
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because by recognizing the geometric series and using the properties of complex numbers, the real part of z can be found to have an absolute value of 1/2.
104 Easy
104 of 979
Analyzing +4 | -1
−1−1−1−...∞  \sqrt { - 1 - \sqrt {1 - \sqrt {1 - ...\infty } } \;} is equal to
View answer and explanation
Correct option: C
ω2{\omega ^2}
Option C (ω2{\omega ^2}) is correct because solving the given equation leads to complex roots, which are cube roots of unity.
105 Easy
105 of 979
Understanding +4 | -1
The additive inverse of 1−i1 - i is
View answer and explanation
Correct option: B
−1+i- 1 + i
Option B (-1 + i) is correct because it is the additive inverse that, when added to the original number, results in zero.

Explore Other Chapters in MATHEMATICS

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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
106 Easy
106 of 979
Application +4 | -1
If ω(≠1)\omega \left( { \ne 1} \right) be a cube root of unity and (1+ω)7=A+Bω,{\left( {1 + \omega } \right)^7} = A + B\omega , then AA and BB are respectively the numbers:
View answer and explanation
Correct option: B
1,  11,\;1
Option B (1, 1) is correct because simplifying the expression using properties of cube roots of unity leads to A = 1 and B = 1.
107 Easy
107 of 979
Application +4 | -1
If ∣6i−3i143i−1203i∣=x+iy,\left| {\begin{array}{cccccccccccccccccccc} {6i}&{ - 3i}&1\\ 4&{3i}&{ - 1}\\ {20}&3&i \end{array}} \right| = x + iy, 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 the determinant is zero after applying column operations.
108 Medium
108 of 979
Analyzing +4 | -1
If a=cos⁡α+isin⁡α,    b=cos⁡β+isin⁡β, c=cos⁡γ+isin⁡γa = \cos \alpha + i\sin \alpha ,\;\;b = \cos \beta + i\sin \beta ,\,c = \cos \gamma + i\sin \gamma and b/c+c/a+a/b=1b/c + c/a + a/b = 1, then cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because by equating the real parts of the given equation involving complex numbers, we directly obtain the desired expression.
109 Easy
109 of 979
Application +4 | -1
If zz and ww are two complex numbers such that ∣zw∣=1\left| {zw} \right| = 1 and arg (z) - arg (w)=π2(w) = \frac{\pi }{2}, then
View answer and explanation
Correct option: A
zˉw=−i\bar zw = - i
Option A (zˉw=−i\bar zw = - i) is correct because, given the conditions on the magnitudes and arguments of z and w, the expression simplifies to -i.
110 Medium
110 of 979
Application +4 | -1
If elog⁡2∣z∣2−∣z∣+4∣z∣2+1=log⁡2(16),e^{\log 2^{\frac{|z|^2-|z|+4}{|z|^2+1}}}=\log _{\sqrt{2}} (16), then ∣z∣=|z| =
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (12\frac{1}{2}) is correct because solving the equation elog⁡2∣z∣2−∣z∣+4∣z∣2+1=log⁡2(16)e^{\log 2^{\frac{|z|^2-|z|+4}{|z|^2+1}}}=\log _{\sqrt{2}} (16) leads to ∣z∣=12|z| = \frac{1}{2}.
111 Medium
111 of 979
Analyzing +4 | -1
If α≠1\mathrm{\alpha } \ne 1 is any nthnth root of unity, then S=1+3α+5α2…S = 1 + 3\alpha + 5{\alpha ^2} \ldots upon nn terms, is equal to
View answer and explanation
Correct option: B
−2n1−α- \frac{{2n}}{{1 - \alpha }}
Option B (−2n1−α- \frac{{2n}}{{1 - \alpha }}) is correct because it is derived by multiplying the original series by alpha, subtracting, and using the properties of the nth root of unity.
112 Easy
112 of 979
Application +4 | -1
The area of the triangle formed by the points representing −z,  iz- z,\;iz and z−izz - iz in the Argand plane is
View answer and explanation
Correct option: C
32∣z∣2\frac{3}{2}{\left| z \right|^2}
The correct answer is C (32∣z∣2\frac{3}{2}{\left| z \right|^2}) because the area of the triangle formed by the given points in the Argand plane is 32\frac{3}{2} times the square of the modulus of z.
113 Medium
113 of 979
Application +4 | -1
If (cos⁡θ+isin⁡θ)(cos⁡2θ+isin⁡2θ)…(cos⁡nθ+isin⁡nθ)=1\left( {\cos \theta + i\sin \theta } \right)\left( {\cos 2\theta + i\sin 2\theta } \right) \ldots \left( {\cos n\theta + i\sin n\theta } \right) = 1 then the value of θ\theta is
View answer and explanation
Correct option: C
4mπn(n+1)\frac{{4m\pi }}{{n\left( {n + 1} \right)}}
The correct answer is C (4mπn(n+1)\frac{{4m\pi }}{{n\left( {n + 1} \right)}}) because using De Moivre's theorem and equating to 1 leads to this solution for θ\theta.
114 Easy
114 of 979
Application +4 | -1
The region represented by {z=x+iy∈C:∣z∣−Re(z)≤1}\{ z = x + iy \in C:\left| z \right| - {\mathop\mathrm{Re}\nolimits} (z) \le 1\} is also given by the inequality
View answer and explanation
Correct option: B
y2≤2(x+12){y^2} \le 2(x + \frac{1}{2})
Option B (y2≤2(x+12){y^2} \le 2(x + \frac{1}{2})) is correct because it represents the inequality derived from the given condition ∣z∣−Re(z)≤1|z| - Re(z) \le 1 after substituting z=x+iyz = x + iy and simplifying.
115 Easy
115 of 979
Application +4 | -1
The equation ∣z−i∣=∣z−1∣,i=−1,\left| {z - i} \right| = \left| {z - 1} \right|,i = \sqrt { - 1} , represents
View answer and explanation
Correct option: B
The line passing through the origin with slope 1
Option B (The line passing through the origin with slope 1) is correct because the equation simplifies to y = x, which represents a line through the origin with a slope of 1.
116 Easy
116 of 979
Analyzing +4 | -1
If the roots of the equation 1x+p+1x+q=1r,  (x≠−p,  x≠−q,  r≠0)\frac{1}{{x + p}} + \frac{1}{{x + q}} = \frac{1}{r},\;\left( {x \ne - p,\;x \ne - q,\;r \ne 0} \right) are equal in magnitude but opposite in sign, then p+qp+q is equal to
View answer and explanation
Correct option: B
2r2r
Option B (2r) is correct because when the roots are equal in magnitude but opposite in sign, the coefficient of x in the quadratic equation derived from the given equation must be zero, leading to p + q = 2r.
117 Easy
117 of 979
Application +4 | -1
If z=3+iz = \sqrt 3 + i, then the argument of z2ez−i{z^2}{e^{z - i}} is equal to
View answer and explanation
Correct option: D
π/3\pi /3
Option D (π/3\pi /3) is correct because the argument of the complex number z2ez−i{z^2}{e^{z - i}} simplifies to π3\frac{\pi}{3} when z=3+iz = \sqrt{3} + i.
118 Easy
118 of 979
Application +4 | -1
If x=3+ix = 3 + i, then x3−3x2−8x+15{x^3} - 3{x^2} - 8x + 15 is equal to
View answer and explanation
Correct option: B
−15- 15
Option A (3) is correct because for a quadratic equation to have equal roots, the discriminant must be zero, and solving for 'a' gives a = 3.
119 Easy
119 of 979
Analyzing +4 | -1
If ∣z−2∣=min⁡{∣z−1∣,  ∣z−5∣},    \left| {z - 2} \right| = \min \left\{ {\left| {z - 1} \right|,\;\left| {z - 5} \right|} \right\},\;\; where zz is a complex number, then
View answer and explanation
Correct option: C
Re(z)∈{32,72}{\mathop\mathrm{Re}\nolimits} \left( z \right) \in \left\{ {\frac{3}{2},\frac{7}{2}} \right\}
Option C is correct because the real part of z can be either 3/2 or 7/2, satisfying the given condition based on the minimum distance.
120 Easy
120 of 979
Application +4 | -1
If ∏p=1reipθ=1,\prod\limits_{p = 1}^r {{e^{ip\theta }} = 1,} where ∏\prod denotes the continued product and i=−1i=\sqrt{-1}, the most general value of θ\theta is
(where, nn is an integer)
View answer and explanation
Correct option: D
4nπr(r+1),n∈I\frac{4 n \pi}{r(r+1)}, n \in I
Option D (4nπr(r+1),n∈I\frac{{4 n \pi }}{{r(r + 1)}}, n \in I) is correct because it correctly sums the geometric series in the exponent and solves for θ\theta using the properties of complex exponentials.
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