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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Medium

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Showing 121-150 of 337 questions

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121 Medium
121 of 337
Application +4 | -1
If ∣ω∣=1|\omega|=1, then the set of points z=ω+1ωz=\omega+\frac{1}{\omega} is satisfies
View answer and explanation
Correct option: A
∣Re⁡(z)∣≤2|\operatorname{Re}(z)| \leq 2
Option A (∣Re⁡(z)∣≤2|\operatorname{Re}(z)| \leq 2) is correct because zz can be expressed as 2Re⁡(ω)2\operatorname{Re}(\omega), and since ∣ω∣=1|\omega|=1, the real part of zz is bounded by 2.
122 Medium
122 of 337
Analyzing +4 | -1
If 2x2x2+5x+2>1x+1\frac{{2x}}{{2{x^2} + 5x + 2}} > \frac{1}{{x + 1}}, then
View answer and explanation
Correct option: C
−2<x<−1- 2 < x < - 1
Option C (−2<x<−1- 2 < x < - 1) is correct because it represents the interval where the inequality holds true after simplifying and analyzing the sign changes.
123 Medium
123 of 337
Application +4 | -1
The perpendicular distance from (1,4,−2)(1,4,-2) to the line joining (2,1,−2)(2,1,-2) and (0,−5,1)(0,-5,1) is
View answer and explanation
Correct option: B
32673 \frac{\sqrt{26}}{7}
Option B (32673 \frac{\sqrt{26}}{7}) is correct because it represents the perpendicular distance calculated using vector cross products and area of the triangle.
124 Medium
124 of 337
Analyzing +4 | -1
For positive integers n1,  n2{n_1},\;{n_2} the value of the expression (1+i)n1+(1+i3)n1+(1+i5)n2+(1+i7)n2,  i=−1  {\left( {1 + i} \right)^{{n_1}}} + {\left( {1 + {i^3}} \right)^{{n_1}}} + {\left( {1 + {i^5}} \right)^{{n_2}}} + {\left( {1 + {i^7}} \right)^{{n_2}}},\;i = \sqrt { - 1} \; is a real number if and only if
View answer and explanation
Correct option: D
n1>0,  n2>0{n_1} > 0,\;{n_2} > 0
Option D (n1 > 0, n2 > 0) is correct because the expression simplifies to a real number only if n1 and n2 are positive integers.
125 Medium
125 of 337
Application +4 | -1
The value 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 inside the modulus leads to ∣2i∣=2|2i| = 2.
126 Medium
126 of 337
Analyzing +4 | -1
Let z=1+\mathrm{z}=1+ ai be a complex number, a>0a>0, such that z3\mathrm{z}^{3} is areal number. Then the sum 1+z+z2+…+z111+\mathrm{z}+\mathrm{z}^{2}+\ldots+\mathrm{z}^{11} is equal to:
View answer and explanation
Correct option: B
−13653i-1365 \sqrt{3} \mathrm{i}
Option B (−13653i-1365 \sqrt{3} i) is correct because by finding the value of 'a' such that z3z^3 is real and then summing the geometric series, we get −13653i-1365 \sqrt{3} i.
127 Medium
127 of 337
Application +4 | -1
If z1&z2z_1 \& z_2 are two complex numbers satisfying ∣z−4∣=Re⁡(z)|z-4|=\operatorname{Re}(z) and having greatest and least argument respectively, then area of triangle formed by origin, z1&z2z_1 \& z_2 is
View answer and explanation
Correct option: C
16 sq. units
Option C (16 sq. units) is correct because the area of the triangle formed by the origin, z1z_1, and z2z_2 is calculated using the coordinates of z1z_1 and z2z_2 which are derived from the given condition.
128 Medium
128 of 337
Application +4 | -1
The range of real number α\alpha for which the equation z+α∣z−1∣+2i=0z+\alpha|z-1|+2 i=0 has a solution, is
View answer and explanation
Correct option: A
[−52,52]\left[-\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{2}\right]
Option A [−52,52]\left[-\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{2}\right] is correct because it represents the range of α\alpha for which the quadratic equation has real solutions.
129 Medium
129 of 337
Application +4 | -1
If complex number zz satisfies ∣z−6i∣=Im⁡(z)|z-6 i|=\operatorname{Im}(z), then range of (arg⁡z−arg⁡zˉ)(\arg z-\arg \bar{z}) will be
View answer and explanation
Correct option: A
[π2,3π2]\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]
Option A ([π2,3π2\frac{\pi}{2}, \frac{3 \pi}{2}]) is correct because the given condition represents a parabola, and analyzing the range of the argument of z leads to the specified interval.
130 Medium
130 of 337
Application +4 | -1
Let aa be a complex number such that ∣a∣<1\left| a \right| < 1 and z1,  z2,…{z_1},\;{z_2}, \ldots be vertices of a polygon such that zk=1+a+a2+...+ak−1{z_k} = 1 + a + {a^2} + ... + {a^{k - 1}}. Then the vertices of the polygon lie within a circle is
View answer and explanation
Correct option: C
∣z−11−a∣=1∣1−a∣\left| {z - \frac{1}{{1 - a}}} \right| = \frac{1}{{\left| {1 - a} \right|}}
Option C (∣z−11−a∣=1∣1−a∣\left| {z - \frac{1}{{1 - a}}} \right| = \frac{1}{{\left| {1 - a} \right|}}) is correct because the vertices lie within a circle centered at 1/(1-a) with radius 1/|1-a|.
131 Medium
131 of 337
Analyzing +4 | -1
Consider an ellipse having its foci at A(z1)A\left(z_1\right) and B(z2)B\left(z_2\right) in the argand plane. If the eccentricity of the ellipse is ee and it is known that origin is an interior point of the ellipse, then
View answer and explanation
Correct option: B
e∈(0,∣z1−z2∣∣z1∣+∣z2∣)e \in\left(0, \frac{\left|z_1-z_2\right|}{\left|z_1\right|+\left|z_2\right|}\right)
Option B is correct because the eccentricity ee must satisfy the condition e∈(0,∣z1−z2∣∣z1∣+∣z2∣)e \in (0, \frac{{|z_1-z_2|}}{{|z_1|+|z_2|}}) for the origin to be an interior point of the ellipse.
132 Medium
132 of 337
Application +4 | -1
The curve in the complex plane given by the equation Re⁡(1/z)=1/4\operatorname{Re}(1 / z)=1 / 4 is a
View answer and explanation
Correct option: B
a circle with radius 2 and centre at (2,0)(2,0)
Option B (a circle with radius 2 and centre at (2,0)) is correct because the equation simplifies to the standard form of a circle (x-2)² + y² = 4.
133 Medium
133 of 337
Understanding +4 | -1
The equation zzˉ+azˉ+aˉz+b=0,    b∈Rz\bar z + a\bar z + \bar az + b = 0,\;\;b \in R represents a circle, if
View answer and explanation
Correct option: B
∣a∣2>b{\left| a \right|^2} > b
Option A is correct because if a point is equidistant from four other points, those four points must lie on a circle (concyclic).
134 Medium
134 of 337
Application +4 | -1
How many solutions are possible for Re⁡(z)2=0\operatorname{Re}(z)^2=0 and ∣z∣=a2(a>0)?|z|=a \sqrt{2}(a>0)?
View answer and explanation
Correct option: D
4
Option D (4) is correct because there are four distinct solutions that satisfy both the real part condition and the magnitude condition.
135 Medium
135 of 337
Application +4 | -1
if locus of point p(z)p(z) which satisfies arg⁡(3+z+z+∣z∣i)=π4\arg (3+z+z+|z| i)=\frac{\pi}{4} is a part of conic C\mathrm{C} then length of latus rectumof C\mathrm{C}
View answer and explanation
Correct option: D
6
Option D (6) is correct because the length of the latus rectum of the hyperbola is 6, derived from simplifying the given argument condition and converting it to the standard hyperbola form.

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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
136 Medium
136 of 337
Analyzing +4 | -1
4  ,44,  48,  416,  …to   ∞\sqrt {4\;} ,\sqrt[4]{4},\;\sqrt[8]{4},\;\sqrt[{16}]{4},\; \ldots \mathrm{to}\;\,\infty are root of equation
View answer and explanation
Correct option: C
x2−5x+4=0{x^2} - 5x + 4 = 0
Option C (x2−5x+4=0{x^2} - 5x + 4 = 0) is correct because the sum and product of the given roots match the coefficients of this quadratic equation.
137 Medium
137 of 337
Application +4 | -1
The value of λ\lambda, for which the equation 6x2+11xy−10y2+x+3+y+λ=06 x^2+11 x y-10 y^2+x+3+y+\lambda=0 represents a pair of straight lines, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated value of λ\lambda that makes the equation represent a pair of straight lines is 15361\frac{{15}}{{361}}, which is not among the given options.
138 Medium
138 of 337
Application +4 | -1
If ∣z−2z+2∣=π6\left| {\frac{{z - 2}}{{z + 2}}} \right| = \frac{\mathrm{\pi }}{6}, then the locus of z is
View answer and explanation
Correct option: B
A circle
Option B (A circle) is correct because the given equation represents a circle in the complex plane.
139 Medium
139 of 337
Application +4 | -1
If 2 and 6 are the roots of the equation ax2+bx+1=0ax^2+b x+1=0, then the quadratic equation, whose roots are 12a+b\frac{1}{2 a+b} and 16a+b\frac{1}{6 a+b}, is:
View answer and explanation
Correct option: D
x2+8x+12=0x^2+8 x+12=0
Option B (π2\frac{\pi}{2}) is correct because, after evaluating the integral using properties of definite integrals, the value of α\alpha is found to be π2\frac{\pi}{2}.
140 Medium
140 of 337
Analyzing +4 | -1
Given that a,  b∈{0,  1,  2,…,9}a,\;b \in \left\{ {0,\;1,\;2, \ldots ,9} \right\} with a+b≠0a + b \ne 0 and that (a+b10)x=(a10+b100)y=1000{\left( {a + \frac{b}{{10}}} \right)^x} = {\left( {\frac{a}{{10}} + \frac{b}{{100}}} \right)^y} = 1000. Then, 1x−1y\frac{1}{x} - \frac{1}{y} is equal to
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{1}{3}) is correct because by setting a = 0 and b = 1, we can solve for x and y and find that 1x−1y=13\frac{1}{x} - \frac{1}{y} = \frac{1}{3}.
141 Medium
141 of 337
Application +4 | -1
Let α\alpha and β\beta be the roots of the equation x2+x+1=0x^2+x+1=0 the equation whose roots are α19,β7\alpha^{19}, \beta^7 is
View answer and explanation
Correct option: D
x2+x+1=0x^2+x+1=0
Option D (x2+x+1=0x^2+x+1=0) is correct because the roots α19\alpha^{19} and β7\beta^7 simplify to ω\omega and ω2\omega^2 respectively, which are the roots of the original equation.
142 Medium
142 of 337
Application +4 | -1
If (3+i)10=a+ib,{\left( {\sqrt 3 + i} \right)^{10}} = a + ib, then aa and bb are respectively
View answer and explanation
Correct option: C
512  &−5123512\;\& - 512\sqrt 3
Option C (512 & -512√3) is correct. By converting the complex number to polar form and using De Moivre's theorem, the result is 512 - 512√3i.
143 Medium
143 of 337
Application +4 | -1
Let αr=ei2πrn(1≤r≤n)\alpha_r=e^{i \frac{2 \pi r}{n}}(1 \leq r \leq n) be the complex number associated with the point ArA_r on Argand plane, and point BB is (2,0)(2,0). Then the value of BA1⋅BA2⋅BA3…BAnB A_1 \cdot B A_2 \cdot B A_3 \ldots B A_n is equal to
View answer and explanation
Correct option: B
2n−12^n-1
Option B (2n−12^n-1) is correct because by substituting x=2 into the polynomial equation and using the properties of complex numbers, we can find the product of the distances.
144 Medium
144 of 337
Analyzing +4 | -1
The value of 1+∑k=014{cos⁡(2k+1)15π+isin⁡(2k+1)15π}1 + \sum \limits_{k = 0}^{14} \left\{ {\cos \frac{{\left( {2k + 1} \right)}}{{15}}\pi + i\sin \frac{{\left( {2k + 1} \right)}}{{15}}\pi } \right\} is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the sum simplifies to 1 after using the properties of complex numbers and geometric series.
145 Medium
145 of 337
Application +4 | -1
If ω\omega is a complex cube root of unity, then (1−ω+ω2)6+(1−ω2+ω)6={\left( {1 - \omega + {\omega ^2}} \right)^6} + {\left( {1 - {\omega ^2} + \omega } \right)^6} =
View answer and explanation
Correct option: D
128
Option D (128) is correct because using the property of cube roots of unity, the expression simplifies to 64 + 64 = 128.
146 Medium
146 of 337
Application +4 | -1
If a=cos⁡θ+isin⁡θa = \cos \theta + i\sin \theta, then 1+a1−a\frac{{1 + a}}{{1 - a}} is equal to
View answer and explanation
Correct option: A
icot⁡θ2i\cot \frac{\theta }{2}
Option A (icot⁡θ2i\cot \frac{\theta }{2}) is correct because it is the simplified form of the given expression using trigonometric identities.
147 Medium
147 of 337
Analyzing +4 | -1
2+5−6−35+14−65\sqrt {2 + \sqrt 5 - \sqrt {6 - 3\sqrt 5 + \sqrt {14 - 6\sqrt 5 } } } is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because the equation holds for three distinct values of x only when a = 2 and r = 0, making a + r = 2.
148 Medium
148 of 337
Application +4 | -1
If a=cos⁡4π3+isin⁡4π3,a = \cos \frac{{4\pi }}{3} + i\sin \frac{{4\pi }}{3}, then the value of (1+a2)3n{\left( {\frac{{1 + a}}{2}} \right)^{3n}} is
View answer and explanation
Correct option: B
(−1)n23n\frac{{{{\left( { - 1} \right)}^n}}}{{{2^{3n}}}}
Option B ((−1)n23n\frac{{{{\left( { - 1} \right)}^n}}}{{{{2^{3n}}}}}) is correct because it is the simplified form of the given expression using trigonometric identities and De Moivre's theorem.
149 Medium
149 of 337
Application +4 | -1
If (1+z)n=a0+a1z+a2z2+⋯+anzn(1+z)^n=a_0+a_1 z+a_2 z^2+\cdots+a_n z^n, where a0,a1a_0, a_1, a2,…,ana_2, \ldots, a_n, are real, then (a0−a2+a4−a6+⋯ )2+(a1−a3+a5−a7+⋯ )2=\left(a_0-a_2+a_4-a_6+\cdots\right)^2+\left(a_1-a_3+a_5-a_7+\cdots\right)^2=
View answer and explanation
Correct option: A
2n2^n
Option A (2n2^n) is correct because substituting z=i into the binomial expansion and comparing magnitudes leads to this result.
150 Medium
150 of 337
Application +4 | -1
If z=x+iy,xy≠0z=x+i y, x y \neq 0, satisfies the equation z2+izˉ=0z^2+i \bar{z}=0, then ∣z2∣\left|z^2\right| is equal to :
View answer and explanation
Correct option: C
1
The correct answer is C (1) because by taking the modulus of the given equation, we find that |z| = 1, and therefore |z2| = 1.
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