Skip to main content
Back to COMPLEX NUMBERS

Your Progress

0 / 612 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 571-600 of 612 questions

Preparing practice and saved progress…

571 Easy
571 of 612
Analyzing +4 | -1
The number of complex numbers zz such that ∣z−1∣=∣z+1∣=∣z−i∣\left| {z - 1} \right| = \left| {z + 1} \right| = \left| {z - i} \right| equals
View answer and explanation
Correct option: B
1
Option B (1) is correct because the only complex number satisfying the given conditions is z=iz = i.
572 Easy
572 of 612
Understanding +4 | -1
If zz is a complex number, then
View answer and explanation
Correct option: B
∣z2∣=∣z∣2\left|z^2\right|=|z|^2
Option B (|z2|=|z|^2) is correct because the magnitude of the square of a complex number is equal to the square of its magnitude.
573 Easy
573 of 612
Application +4 | -1
If nn is a positive integer, then (1+i)n+(1−i)n{\left( {1 + i} \right)^n} + {\left( {1 - i} \right)^n} is equal to
View answer and explanation
Correct option: C
(2)n+2cos⁡(nπ4){\left( {\sqrt 2 } \right)^{n + 2}}\cos \left( {\frac{{n\pi }}{4}} \right)
Option C ((2)n+2cos⁡(nπ4){\left( {\sqrt 2 } \right)^{n + 2}}\cos \left( {\frac{{n\pi }}{4}} \right)) is correct because it is the simplified form of the given expression using De Moivre's theorem.
574 Easy
574 of 612
Application +4 | -1
If α\alpha is a complex number satisfying the equation α2+α+1=0{\alpha ^2} + \alpha + 1 = 0, then α31{\alpha ^{31}} is equal to
View answer and explanation
Correct option: A
α\alpha
Option A (α\alpha) is correct because α31{\alpha ^{31}} simplifies to α\alpha when α\alpha is a complex number satisfying the given equation.
575 Easy
575 of 612
Analyzing +4 | -1
The locus of the points representing the complex numbers zz for which ∣z∣−2=∣z−i∣−∣z+5  i∣=0\left| z \right| - 2 = \left| {z - i} \right| - \left| {z + 5\;i} \right| = 0 is
View answer and explanation
Correct option: C
The single point (0,  −2)\left( {0,\; - 2} \right)
Option C is correct. The locus of the complex number z is the single point (0, -2) because it satisfies both the circle equation and the perpendicular bisector condition.
576 Easy
576 of 612
Application +4 | -1
If x+iy=a+iba−ibx+i y=\frac{a+i b}{a-i b}, then x2+y2=x^2+y^2=
View answer and explanation
Correct option: A
1
Option A (1) is correct because when x+iy is multiplied by its conjugate, the result is 1, which means x² + y² = 1.
577 Easy
577 of 612
Application +4 | -1
The value of (2−ω)(2−ω2)(2−ω10)(2−ω11)\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right)\left( {2 - {\omega ^{10}}} \right)\left( {2 - {\omega ^{11}}} \right), where ω\omega is the complex cube root of unity, is
View answer and explanation
Correct option: A
49
Option A (49) is correct because by using the property that ω3=1\omega^3 = 1 and 1+ω+ω2=01 + \omega + \omega^2 = 0, the expression simplifies to 49.
578 Easy
578 of 612
Analyzing +4 | -1
Let zz and ww be two complex numbers such that ∣z∣=∣w∣=1|z|=|w|=1 and ∣z+iw∣=∣z−iwˉ∣=2|z+i w|=|z-i \bar{w}|=2. Then zz equals
View answer and explanation
Correct option: C
1 or -1
Option C (1 or -1) is correct because the given conditions imply that z must be a real number with magnitude 1.
579 Easy
579 of 612
Application +4 | -1
The number of complex numbers zz which satisfy z2+2∣z∣2=2z^2+2|z|^2=2 is
View answer and explanation
Correct option: D
4
Option D (4) is correct because by considering both real and imaginary cases for z, we find four complex numbers that satisfy the given equation.
580 Easy
580 of 612
Application +4 | -1
The radius of the circle ∣z−iz+i∣=5\left| {\frac{{z - i}}{{z + i}}} \right| = 5 is given by
View answer and explanation
Correct option: B
512\frac{5}{{12}}
Option B (5/12) is correct because the radius of the circle is calculated using the formula derived from the given complex equation.
581 Easy
581 of 612
Analyzing +4 | -1
If ∣z−2−i∣=∣z∣∣sin⁡(π4−arg⁡(z))|z-2-i|=|z| \mid \sin \left(\frac{\pi}{4}-\arg (z)\right), then locus of zz is
View answer and explanation
Correct option: C
a parabola
Option C (a parabola) is correct because the given equation simplifies to the form (x−2)2+(y−1)2=12∣x−y∣\sqrt{{(x-2)^2 + (y-1)^2}} = \frac{{1}}{{\sqrt{{2}}}} |x-y|, which represents a parabola.
582 Easy
582 of 612
Application +4 | -1
If ω\omega is a cube root of unity, then the value of (1−ω+ω2)5+(1+ω−ω2)5{\left( {1 - \omega + {\omega ^2}} \right)^5} + {\left( {1 + \omega - {\omega ^2}} \right)^5} is
View answer and explanation
Correct option: B
32
Option B (32) is correct because using the properties of cube roots of unity (1 + ω + ω² = 0 and ω³ = 1), the expression simplifies to 32.
583 Easy
583 of 612
Analyzing +4 | -1
If a,b,ca, b, c are integers, not all equal and ω(≠1)\omega(\neq 1) is a cube root of unity, then minimum value of ∣a+bω+cω2∣\left|a+b \omega+c \omega^2\right| is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the minimum value of the expression ∣a+bω+cω2∣\left|a+b \omega+c \omega^2\right| is 1 when a, b, and c are integers not all equal, and ω\omega is a cube root of unity.
584 Easy
584 of 612
Application +4 | -1
The complex number z=x+iyz = x + iy which satisfy the equation ∣z−5iz+5i∣=1\left| {\frac{{z - 5i}}{{z + 5i}}} \right| = 1 lies on
View answer and explanation
Correct option: A
The axis of xx
Option A (The axis of x) is correct because the given condition simplifies to y = 0, which represents the x-axis.
585 Easy
585 of 612
Application +4 | -1
If ∣z−1z∣=1\left|z-\frac{1}{\mathrm{z}}\right|=1, then:
View answer and explanation
Correct option: A
∣z∣max⁡=1+52|z|_{\max }=\frac{1+\sqrt{5}}{2}
Option A (∣z∣max⁡=1+52|z|_{\max } = \frac{1+\sqrt{5}}{2}) is correct because by using the reverse triangle inequality, we can find the maximum value of ∣z∣|z|.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs DIFFERENTIAL EQUATIONS 977 MCQs
586 Easy
586 of 612
Application +4 | -1
The value of (−12+3i2)1000{\left( { - \frac{1}{2} + \frac{{\sqrt 3 i}}{2}} \right)^{1000}} is
View answer and explanation
Correct option: D
ω\omega
Option D (ω\omega) is correct because (−12+3i2)1000{\left( { - \frac{{1}}{{2}} + \frac{{\sqrt 3 i}}{{2}}} \right)^{1000}} simplifies to ω\omega using the properties of complex cube roots of unity.
587 Easy
587 of 612
Analyzing +4 | -1
Let for some real numbers α\alpha and β,  a=α−iβ.\beta ,\,\,a = \alpha - i\beta . If the system of equations 4ix+(1+i)y=04ix + \left( {1 + i} \right)y = 0 and 8(cos⁡2π3+isin⁡2π3)x+aˉy=08\left( {\cos \frac{{2\pi }}{3} + i\sin \frac{{2\pi }}{3}} \right)x + \bar ay = 0 has more than one solution, then αβ\frac{\alpha }{\beta } is equal to
View answer and explanation
Correct option: B
2−32 - \sqrt 3
The correct answer is B (2−32 - \sqrt 3) because it is the value of αβ\frac{\alpha}{\beta} calculated using the condition for the system of equations to have more than one solution.
588 Easy
588 of 612
Analyzing +4 | -1
If ∣z1∣=∣z2∣=…=∣zn∣=1,  \left| {{z_1}} \right| = \left| {{z_2}} \right| = \ldots = \left| {{z_n}} \right| = 1,\; then the value of ∣z1+z2+z3+...+zn∣\left| {{z_1} + {z_2} + {z_3} + ... + {z_n}} \right| is
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 (∣1z1+1z2+...+1zn∣\left| {\frac{1}{{{{z_1}}}} + \frac{1}{{{{z_2}}}} + ... + \frac{1}{{{{z_n}}}}} \right|) is correct because it uses the property that if ∣z∣=1|z| = 1, then zˉ=1z\bar{z} = \frac{1}{z} and ∣z∣=∣zˉ∣|z| = |\bar{z}|.
589 Easy
589 of 612
Analyzing +4 | -1
Number of roots of the equation z10−z5−992=0z^{10}-z^5-992=0, where real parts are negative, is
View answer and explanation
Correct option: C
5
Option C (5) is correct because solving the equation yields two values for z5, one of which (z5 = -31) has 5 solutions with negative real parts.
590 Easy
590 of 612
Application +4 | -1
The mirror image of the curve arg⁡(z−3z−i)=π6\arg \left(\frac{z-3}{z-i}\right)=\frac{\pi}{6}, i=−1i=\sqrt{-1} in the real axis, is
View answer and explanation
Correct option: D
arg⁡(z+iz−3)=π6\arg \left(\frac{z+i}{z-3}\right)=\frac{\pi}{6}
Option D (arg⁡(z+iz−3)=π6\arg \left(\frac{{z + i}}{{z - 3}}\right) = \frac{\pi }{6}) is correct because the mirror image of a complex number z in the real axis is its conjugate zˉ\bar{z}, and using the properties of arguments, the equation transforms to the given answer.
591 Easy
591 of 612
Analyzing +4 | -1
If ∣z∣=1|z|=1 and ω=z−1z+1(\omega=\frac{z-1}{z+1}( where z≠−1)z \neq-1), then Re⁡(ω)\operatorname{Re}(\omega) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the real part of the given expression, where |z|=1 and z is not -1, simplifies to zero, indicating a purely imaginary number.
592 Easy
592 of 612
Remembering +4 | -1
The real part of [1+cos⁡(π5)+isin⁡(π5)]\left[ {1 + \cos \left( {\frac{\pi }{5}} \right) + i\sin \left( {\frac{\pi }{5}} \right)} \right] is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because after rationalizing the denominator, the real part of the expression simplifies to 1/2.
593 Easy
593 of 612
Application +4 | -1
If x+(1/x)=2cos⁡θx+(1 / x)=2 \cos \theta, then the value of xx is
View answer and explanation
Correct option: A
cos⁡θ±isin⁡θ\cos \theta \pm i \sin \theta
Option A (cos⁡θ±isin⁡θ\cos \theta \pm i \sin \theta) is the correct solution because it is derived by solving the quadratic equation resulting from the given equation.
594 Easy
594 of 612
Analyzing +4 | -1
If (1+x+x2)n=a0+a1x+a2x2+…+\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+ a2nx2na_{2 n} x^{2 n}, then value of a0+a3+a6+…a_0+a_3+a_6+\ldots is
View answer and explanation
Correct option: D
3n−13^{n-1}
Option D (3n−13^{n-1}) is correct because by substituting x=1,ω,ω2x=1, \omega, \omega^2 into the given polynomial and adding the resulting equations, we can isolate the desired sum.
595 Easy
595 of 612
Application +4 | -1
For a positive integer nn, the expression (1−i)n(1−1i)n(1-i)^n\left(1-\frac{1}{i}\right)^n equals
View answer and explanation
Correct option: C
2n2^n
Option C (2n2^n) is correct because after simplification, the expression equals 2n2^n.
596 Easy
596 of 612
Application +4 | -1
If ∣z2−1∣=∣z∣2+1\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1, then zz lies on
View answer and explanation
Correct option: B
The imaginary axis
Option B (The imaginary axis) is correct because the given condition implies that the argument of z2z^2 is π\pi, which means the argument of z is π2\frac{\pi}{2}, indicating that z lies on the imaginary axis.
597 Easy
597 of 612
Analyzing +4 | -1
If ω(≠1)\omega(\neq 1) is cube root of unity satisfying 1a+ω+1b+ω+1c+ω=2ω2\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2 \omega^2 and 1a+ω2+1b+ω2+1c+ω2=2ω\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2 \omega, then the value of 1a+1+1b+1+1c+1\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because ω\omega and ω2\omega^2 are roots of the equation 1a+x+1b+x+1c+x=2x\frac{{1}}{{a+x}} + \frac{{1}}{{b+x}} + \frac{{1}}{{c+x}} = \frac{{2}}{{x}}, so substituting x=1x=1 gives 1a+1+1b+1+1c+1=2\frac{{1}}{{a+1}} + \frac{{1}}{{b+1}} + \frac{{1}}{{c+1}} = 2.
598 Easy
598 of 612
Application +4 | -1
The conjugate of the complex number (1+i)21−i\frac{{{{\left( {1 + i} \right)}^2}}}{{1 - i}} is
View answer and explanation
Correct option: D
−1−i- 1 - i
Option D (−1−i- 1 - i) is correct because after simplifying the complex number and taking its conjugate, we get -1-i.
599 Easy
599 of 612
Application +4 | -1
If ω(≠1)\omega \left( { \ne 1} \right) be a cube root of unity and (1+ω2)n=(1+ω4)n{\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n}, then the least positive value of nn is
View answer and explanation
Correct option: B
3
Option B (3) is correct because it is the smallest positive integer n that satisfies the equation ωn=1{\omega}^n = 1, where ω\omega is a cube root of unity.
600 Easy
600 of 612
Application +4 | -1
If zz and ww are two non-zero complex numbers such that ∣zw∣=1\left| {zw} \right| = 1 and arg⁡(z)−arg⁡(w)=π2\arg \left( z \right) - \arg \left( w \right) = \frac{\pi }{2}, then zˉw\bar{z}w is equal to
View answer and explanation
Correct option: D
−i- i
Option D (−i-i) is correct because by expressing z and w in exponential form and using the given conditions, zˉw\bar{z}w simplifies to −i-i.
Revision notes and key takeaways

Review COMPLEX NUMBERS in your textbook or class notes, then use this set to check your understanding.

Showing 571-600 of 612 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - COMPLEX NUMBERS MCQs (All) 979 MCQs → ENGINEERING MATHEMATICS - COMPLEX NUMBERS MCQs (Hard) 30 MCQs → ENGINEERING MATHEMATICS - COMPLEX NUMBERS MCQs (Medium) 337 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs

Your practice

0 / 612 answered · 612 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →