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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Medium

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Showing 31-60 of 337 questions
31 Medium
31 of 337
Understanding +4 | -1
The region of argand diagram defined by ∣z−1∣+∣z+1∣≤4|z-1|+|z+1| \leq 4 is
View answer and explanation
Correct option: C
interior and boundary of an ellipse
Option C (interior and boundary of an ellipse) is correct because the given inequality defines the set of points whose sum of distances from two fixed points is less than or equal to a constant, which is the definition of an ellipse's interior and boundary.
32 Medium
32 of 337
Application +4 | -1
If (1+i)(1+2i)(1+3i)…(1+ni)=x+iy(1+i)(1+2 i)(1+3 i) \ldots(1+n i)=x+i y, then 2.5.10…(1+n2)2.5 .10 \ldots\left(1+n^2\right) is equal to
View answer and explanation
Correct option: C
x2+y2x^2+y^2
Option C (x2+y2x^2+y^2) is correct because multiplying the given equation by its conjugate results in the product of magnitudes, which equals x2+y2x^2 + y^2.
33 Medium
33 of 337
Application +4 | -1
If zz is a complex number in the Argand plane such that arg⁡(z−33z+33)=π3\arg \left( {\frac{{z - 3\sqrt 3 }}{{z + 3\sqrt 3 }}} \right) = \frac{\pi }{3} then the lous of zz is
View answer and explanation
Correct option: B
∣z−3i∣=6,  Im  (z)>0\left| {z - 3i} \right| = 6,\;\mathrm{Im\;}\left( z \right) > 0
Option B (∣z−3i∣=6,  Im  (z)>0\left| {z - 3i} \right| = 6,\;\mathrm{Im\;}\left( z \right) > 0) is correct because the locus represents a circle with center (0, 3) and radius 6, but only the part above the x-axis is considered due to the argument condition.
34 Medium
34 of 337
Application +4 | -1
If ω=cos⁡πn+isin⁡πn\omega=\cos \frac{\pi}{n}+i \sin \frac{\pi}{n}, then value of 1+ω+ω2+…+ωn−11+\omega+\omega^2+\ldots+\omega^{n-1} is
View answer and explanation
Correct option: C
1+icot⁡(π/2n)1+i \cot (\pi / 2 n)
Option C (1+icot⁡(π/2n)1+i \cot (\pi / 2 n)) is correct because it represents the sum of the geometric series with complex terms, simplified using trigonometric identities.
35 Medium
35 of 337
Application +4 | -1
The complex numbers zz having positive argument and satisfying ∣2−3i∣<∣z∣\left| {2 - 3i} \right| < \left| z \right|, is
View answer and explanation
Correct option: A
125+165i\frac{{12}}{5} + \frac{{16}}{5}i
Option A (125+165i\frac{{12}}{5} + \frac{{16}}{5}i) is correct because it satisfies both the positive argument condition and the inequality |2 - 3i| < |z|.
36 Medium
36 of 337
Application +4 | -1
If 1,  ω,  ω21,\;\omega ,\;{\omega ^2} are the cube roots of unity, then (1−ω+ω2)(1−ω2+ω4)(1−ω4+ω8)(1−ω8+ω16)\left( {1 - \omega + {\omega ^2}} \right)\left( {1 - {\omega ^2} + {\omega ^4}} \right)\left( {1 - {\omega ^4} + {\omega ^8}} \right)\left( {1 - {\omega ^8} + {\omega ^{16}}} \right) upto 2n2n factors is
View answer and explanation
Correct option: B
22n{2^{2n}}
Option B (22n{2^{2n}}) is correct because each pair of factors simplifies to 4, and there are n such pairs.
37 Medium
37 of 337
Application +4 | -1
If w=cos⁡(π/n)+isin⁡(π/n)w=\cos (\pi / n)+i \sin (\pi / n), then value of 1+w+1+w+ w2+⋯+wn−1w^2+\cdots+w^{n-1} is
View answer and explanation
Correct option: C
1+icot⁡(π/2n)1+i \cot (\pi / 2 n)
Option C (1+icot⁡(π/2n)1+i \cot (\pi / 2 n)) is correct because after simplifying the geometric series and using trigonometric identities, the expression equals 1+icot⁡(π/2n)1+i \cot (\pi / 2 n).
38 Medium
38 of 337
Understanding +4 | -1
The set C={z:zzˉ+azˉ+aˉz+b=0,  b∈RC = \{ z:z\bar z + a\bar z + \bar az + b = 0,\;b \in R and b<∣a∣2}b < {\left| a \right|^2}\} is
View answer and explanation
Correct option: B
An infinite set
Option B is correct because the given equation represents a circle in the complex plane, and a circle has an infinite number of points.
39 Medium
39 of 337
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.
40 Medium
40 of 337
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.
41 Medium
41 of 337
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.
42 Medium
42 of 337
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.
43 Medium
43 of 337
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.
44 Medium
44 of 337
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}.
45 Medium
45 of 337
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.
46 Medium
46 of 337
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.
47 Medium
47 of 337
Application +4 | -1
Let x=α+β,  y=αω+βω2,  z=αω2+βω,  ωx = \alpha + \beta ,\;y = \alpha \omega + \beta {\omega ^2},\;z = \alpha {\omega ^2} + \beta \omega ,\;\omega is an imaginary cube root of unity. The value of xyzxyz is
View answer and explanation
Correct option: C
α3+β3{\alpha ^3} + {\beta ^3}
Option C (α3+β3{\alpha ^3} + {\beta ^3}) is correct because substituting the given expressions for x,y,zx, y, z and using the properties of cube roots of unity simplifies the product xyzxyz to α3+β3\alpha^3 + \beta^3.
48 Medium
48 of 337
Application +4 | -1
If 1,ω,ω2,…,ωn−11, \omega, \omega^2, \ldots, \omega^{n-1} are the n,nn, nth roots of unity, then (7−ω)(7−ω2)⋯(7−ωn−1)(7-\omega)\left(7-\omega^2\right) \cdots\left(7-\omega^{n-1}\right) is equal to
View answer and explanation
Correct option: C
7n−16\frac{7^n-1}{6}
Option C (7n−16\frac{{7^n-1}}{{6}}) is correct because by using the factorization of xn−1x^n - 1 and substituting x=7x=7, we can find the value of the given expression.
49 Medium
49 of 337
Application +4 | -1
If w=z−(1/5)izw=\frac{z-(1 / 5) i}{z} and ∣w∣=1|w|=1, then complex number zz lies on
View answer and explanation
Correct option: C
a line
Option C (a line) is correct because the condition |w| = 1 leads to an equation that represents a straight line in the complex plane.
50 Medium
50 of 337
Application +4 | -1
If z1,z2,z3z_1, z_2, z_3 are complex number such that ∣z1∣=2,∣z2∣=3,∣z3∣=4,\left|z_1\right|=2,\left|z_2\right|=3,\left|z_3\right|= 4 , then maximum value of ∣z1−z2∣2+∣z2−z3∣2+∣z3−z1∣2\left|z_1-z_2\right|^2+\left|z_2-z_3\right|^2+\left|z_3-z_1\right|^2 is
View answer and explanation
Correct option: C
87
Option C (87) is correct because the maximum value of the expression is obtained by minimizing the negative terms, which leads to the result 87.
51 Medium
51 of 337
Application +4 | -1
The three angular points of a triangle are given by Z=α,Z=β,Z=γZ=\alpha, Z=\beta, Z=\gamma, where α,β,γ\alpha, \beta, \gamma are complex numbers, then the perpendicular from the angular point Z=αZ=\alpha to opposite side is given by the equation :
View answer and explanation
Correct option: B
Re⁡(Z−αβ−γ)=0\operatorname{Re}\left(\frac{Z-\alpha}{\beta-\gamma}\right)=0
Option B (Re(Z−αβ−γ\frac{{Z-\alpha}}{{\beta-\gamma}})=0) is correct because the perpendicularity condition implies that the argument of the complex number (Z-α)/(β-γ) is π/2, making its real part zero.
52 Medium
52 of 337
Application +4 | -1
If ∣z∣=5|z|=5, then the points representing the complex number −i+15z-i+\frac{15}{z} lies on the circle
View answer and explanation
Correct option: B
whose centre is (0,−1)(0,-1) and radius =3=3
Option B (whose centre is (0,-1) and radius =3) is correct because it correctly identifies the center and radius of the circle on which the transformed complex number lies.
53 Medium
53 of 337
Application +4 | -1
Statement-I: Roots of quadratic equation x2+3x+5=0x^2+3 x+5=0 is x=−3±i112x=\frac{-3 \pm {i} \sqrt{11}}{2}.

Statement-II: If x2−x+2=0x^2-x+2=0 is a quadratic equation, then its roots are 1±i72\frac{1 \pm {i} \sqrt{7}}{2}.
View answer and explanation
Correct option: C
Both are correct
Option C (Both are correct) is correct because both statements correctly calculate the roots of the given quadratic equations using the quadratic formula.
54 Medium
54 of 337
Application +4 | -1
If (cos⁡θ+isin⁡θ)  (cos⁡2θ+isin⁡2θ)…  (cos⁡nθ+isin⁡nθ)=1(\cos \mathrm{\theta } + i\sin \mathrm{\theta })\mathrm{\;}(\cos 2\mathrm{\theta } + i\sin 2\mathrm{\theta }) \ldots \;(\cos n\mathrm{\theta } + i\sin n\mathrm{\theta }) = 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)}}
Option C 4mπn(n+1)\frac{{4m\pi }}{{n\left( {n + 1} \right)}} is correct because it is derived by using De Moivre's theorem and equating the imaginary part to zero to find the general solution for θ\theta.
55 Medium
55 of 337
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
Option B (1) is correct because by substituting p=cot⁡θp = \cot \theta and simplifying using trigonometric identities and Euler's formula, the expression equals 1.
56 Medium
56 of 337
Application +4 | -1
If 1−iα1+iα=A+iB\frac{1-i \alpha}{1+i \alpha}=A+i B, then A2+B2A^2+B^2 equals
View answer and explanation
Correct option: A
1
Option A (1) is correct because the magnitude of a complex number divided by its conjugate is always 1.
57 Medium
57 of 337
Application +4 | -1
The number of values of θ∈(0,π]\theta \in(0, \pi], such that ( cos⁡θ+\cos \theta+ isin⁡θ)(cos⁡3θ+isin⁡3θ)(cos⁡5θ+isin⁡5θ)(cos⁡7θi \sin \theta)(\cos 3 \theta+i \sin 3 \theta)(\cos 5 \theta+i \sin 5 \theta)(\cos 7 \theta +isin⁡7θ)(cos⁡9θ+isin⁡9θ)=−1+i \sin 7 \theta)(\cos 9 \theta+i \sin 9 \theta)=-1 is
View answer and explanation
Correct option: B
13
Option B (13) is correct because solving the equation cos(25θ) = -1 within the given interval (0, π] yields 13 possible values for θ.
58 Medium
58 of 337
Application +4 | -1
If ∣z+4∣≤3\left| {z + 4} \right| \le 3, then the greatest and the least value of ∣z+1∣\left| {z + 1} \right| are respectively
View answer and explanation
Correct option: B
6,06,0
Option B (6,06,0) is correct because these are the greatest and least values of ∣z+1∣|z+1| based on the given inequality and geometric interpretation.
59 Medium
59 of 337
Application +4 | -1
ei(2kcot⁡−1m)(mi+1mi−1)k=……e^{i\left(2 k \operatorname{cot}^{-1} m\right)}\left(\frac{m i+1}{m i-1}\right)^k=\ldots \ldots where k∈Z,m>0k \in Z, m>0
View answer and explanation
Correct option: B
1
Option B (1) is correct because the expression simplifies to 1 using Euler's formula and trigonometric identities.
60 Medium
60 of 337
Application +4 | -1
If the complex numbers z1=a+i,  z2=1++ib,z3=0{z_1} = a + i,\;{z_2} = 1 + + ib,{z_3} = 0 form the vertices of equilateral triangle (a,  ba,\;b are real numbers between 0 and 1), then
View answer and explanation
Correct option: B
a=2−3,  b=2−3a = 2 - \sqrt 3 ,\;b = 2 - \sqrt 3
Option B (a=2−3,  b=2−3a = 2 - \sqrt 3 ,\;b = 2 - \sqrt 3) is correct because by using the properties of equilateral triangles and complex numbers, we can derive the values of a and b.
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