If $z_1$ lies on $|z|=r$, then equation of tangent at $z_1$ is

$\frac{z}{\bar{z}_1}+\frac{\bar{z}}{z_1}=r$

$\frac{z}{\bar{z}_1}+\frac{\bar{z}}{z_1}=2$

$\frac{z}{z_1}+\frac{\bar{z}}{\bar{z}_1}=r$

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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Easy

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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.

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Showing 451-480 of 612 questions

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451 Easy

451 of 612

Application +4 | -1

The argument of the complex number

\frac{{13 - 5i}}{{4 - 9i}}

View answer and explanation

Correct option: B

\frac{\pi }{4}

Option B

\frac{\pi }{4}

is correct because it is the argument of the simplified complex number, found by rationalizing the denominator and calculating the arctangent of the imaginary to real part ratio.

452 Easy

452 of 612

Application +4 | -1

\left|z_1-1\right|<1,\left|z_2-2\right|<2

and

\left|z_3-3\right|<3

, then

\mid z_1+

z_2+z_3 \mid

View answer and explanation

Correct option: C

is less than 12

Option C (is less than 12) is correct because by triangle inequality, the magnitude of the sum is less than the sum of the magnitudes.

453 Easy

453 of 612

Analyzing +4 | -1

z \in \mathbf{C}, z \notin \mathbf{R}

, and

a=z^2+3 z+5

, then

a

cannot take value

View answer and explanation

Correct option: C

11 / 4

Option C (11/4) is correct because the expression a = z² + 3z + 5 cannot take the value 11/4 when z is not a real number.

454 Easy

454 of 612

Analyzing +4 | -1

The value of

\sum\limits_{{k=1}}^{n-1}(n-k) \cos \frac{2 k \pi}{n} \quad(n \geq 3

is an integer)

View answer and explanation

Correct option: A

-\frac{n}{2}

Option A (-n/2) is correct because the sum can be simplified using trigonometric series identities.

455 Easy

455 of 612

Application +4 | -1

\omega

is a non-real cube root of unity, then the value of

\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}+\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}

is equal to:

View answer and explanation

Correct option: D

-1

Option D (-1) is correct because by multiplying the numerator and denominator of each fraction by appropriate powers of

\omega

, the expression simplifies to

\omega + \frac{{1}}{{\omega}} = -1

456 Easy

456 of 612

Analyzing +4 | -1

Let

S=\left\{z=x+i y: \frac{2 z-3 i}{4 z+2 i}\right.

is a real number

\}

. Then which of the following is NOT correct?

View answer and explanation

Correct option: B

(x, y)=\left(0,-\frac{1}{2}\right)

Option B

((x, y) = (0, -\frac{{1}}{{2}}))

is correct because when

x = 0

and

y = -\frac{{1}}{{2}}

, the denominator of the expression becomes zero, making the expression undefined, thus it is not a real number.

457 Easy

457 of 612

Application +4 | -1

z=(3+7 i)(\lambda+i \mu)

, when

\lambda, \mu \in I \sim\{0\}

and

i=\sqrt{-1}

, is purely imaginary then minimum value of

|z|^2

View answer and explanation

Correct option: D

3364

Option D (3364) is correct because it is the minimum value of

|z|^2

when z is purely imaginary, calculated using the given conditions.

458 Easy

458 of 612

Understanding +4 | -1

The value of

1 + {i^2} + {i^4} + {i^6} + \ldots + {i^{2n}}

View answer and explanation

Correct option: D

Cannot be determined

Option D (Cannot be determined) is correct because the sum depends on the value of

n

, which is not specified.

459 Easy

459 of 612

Application +4 | -1

The smallest positive integer

n

for which

[(1+i) /

(1-i)]^n=1

View answer and explanation

Correct option: B

Option B (4) is correct because the expression simplifies to i, and i raised to the power of 4 equals 1, which is the smallest positive integer that satisfies the equation.

460 Easy

460 of 612

Application +4 | -1

z_1

lies on

|z|=r

, then equation of tangent at

z_1

View answer and explanation

Correct option: A

\frac{z}{z_1}+\frac{\bar{z}}{\bar{z}_1}=2

Option A (

\frac{{z}}{{z_1}}+\frac{{\bar{z}}}{{\bar{z}_1}}=2

) is correct because it represents the equation of the tangent to the circle

|z|=r

at the point

z_1

461 Easy

461 of 612

Application +4 | -1

a+i b=\sum\limits_{{k=1}}^{101} i^k

, then

(a, b)

equals

View answer and explanation

Correct option: A

(0,1)

Option A (

(0,1)

) is correct because the sum of the series

\sum_{k=1}^{101} i^k

simplifies to

i

, which corresponds to the complex number

0 + 1i

462 Easy

462 of 612

Analyzing +4 | -1

The points

A, B

and

C

represent the complex numbers

z_1, z_2,(1-i) z_1+i z_2

respectively, on the complex plane (where,

i=\sqrt{-1}

). The

\triangle A B C

, is

View answer and explanation

Correct option: C

isosceles and right angled

Option C (isosceles and right angled) is correct because the triangle satisfies both the conditions for being isosceles (AB=CA) and right-angled ((AB)²⁺(CA)²=(BC)²).

463 Easy

463 of 612

Analyzing +4 | -1

b

and

c

are odd integers, then the equation

{x^2} + bx + c = 0

has

View answer and explanation

Correct option: C

No integer roots

Option C (No integer roots) is correct because if b and c are odd, the discriminant is not a perfect square, thus the roots are irrational.

464 Easy

464 of 612

Application +4 | -1

z=\frac{1+\cos \theta+i \sin \theta}{\sin \theta+i(1+\cos \theta)}(0 < \theta < \pi / 2)

then

|z|

equals

View answer and explanation

Correct option: C

Option C (1) is correct because by using the property |z1/z2| = |z1|/|z2|, the magnitude of the numerator and denominator are equal, resulting in |z| = 1.

465 Easy

465 of 612

Understanding +4 | -1

z_1, z_2

and

z_3

are three complex numbers in AP, then they lie on

View answer and explanation

Correct option: B

a straight line

Option B (a straight line) is correct because if complex numbers are in AP, the middle term is the average of the other two, implying collinearity.

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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

466 Easy

466 of 612

Analyzing +4 | -1

If at least one value of the complex number

z = x + i\;y

satisfy the condition

\left| {z + \sqrt 2 } \right| = \sqrt {{a^2} - 3a + 2}

and the inequality

\left| {z + i\sqrt 2 } \right| < a,

then

View answer and explanation

Correct option: A

a > 2

Option A is correct. The condition for the intersection of the two circles leads to the inequality

a > 2

467 Easy

467 of 612

Analyzing +4 | -1

A complex number

z

is said to be unimodular if

|z|=1

.
Suppose

z_1

and

z_2

are complex numbers such that

\frac{z_1-2 z_2}{2-z_1 \bar{z}_2}

is unimodular and

z_2

is not unimodular. Then the point

z_1

lies on a

View answer and explanation

Correct option: A

circle of radius

z

The point

z_1

lies on a circle of radius 2 because the unimodular condition simplifies to

|z_1| = 2

468 Easy

468 of 612

Analyzing +4 | -1

\alpha(\neq 1)

is a fifth root of unity and

\beta(\neq 1)

is a fourth root of unity then

z=(1+\alpha)(1+\beta)\left(1+\alpha^2\right)\left(1+\beta^2\right)\left(1+\alpha^3\right)\left(1+\beta^3\right)

equals

View answer and explanation

Correct option: D

Option D (0) is correct because the expression (1+

\beta

)(1+

\beta^2

) equals zero since

\beta

is a fourth root of unity (

\neq

1).

469 Easy

469 of 612

Application +4 | -1

A value of

\theta

for which

\mathrm{Z}=\frac{2+3 i \sin \theta}{1-2 i \sin \theta}

is purely imaginary is

View answer and explanation

Correct option: A

\sin ^{-1} \frac{1}{\sqrt{3}}

Option A (

\sin^{-1} \frac{1}{\sqrt{3}}

) is correct because it makes the real part of the complex number Z equal to zero, resulting in a purely imaginary number.

470 Easy

470 of 612

Application +4 | -1

x+i y=\sqrt{\frac{3+i}{1+3 i}}

then

\left(x^2+y^2\right)^2

equals

View answer and explanation

Correct option: D

Option D (1) is correct because by simplifying the expression inside the square root and then calculating

(x^2+y^2)^2

, it can be shown that the result is 1.

471 Easy

471 of 612

Application +4 | -1

z

is a complex number and

i=\sqrt{-1}

, then the minimum possible value of

|z|^2+|z-3|^2+|z-6 i|^2

View answer and explanation

Correct option: B

Option B (30) is correct because the minimum value is found by completing the square and recognizing the minimum occurs when z = 1 + 2i.

472 Easy

472 of 612

Application +4 | -1

The locus of the complex number

z

in an Argand plane satisfying the equation

\arg (z+i)-\arg (z-i)=(\pi / 2)

View answer and explanation

Correct option: A

The boundary of a circle

Option A (The boundary of a circle) is correct because the locus satisfying the given condition represents a circle.

473 Easy

473 of 612

Application +4 | -1

Let

z=\cos \theta+i \sin \theta

, where

i=\sqrt{-1}

. Then the value of

\sum\limits_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right)

\theta=2^{\circ}

View answer and explanation

Correct option: D

\frac{1}{4 \sin 2^{\circ}}

Option D (

\frac{{1}}{{4 \sin 2^{\circ}}}

) is correct because the sum of the imaginary parts simplifies to this expression using trigonometric identities.

474 Easy

474 of 612

Analyzing +4 | -1

P,\;Q,\;R,\;S

are represented by the complex numbers

4 + i,\;1 + 6i,\; - 4 + 3i,\; - 1 - 2i

respectively, then

PQRS

is a

View answer and explanation

Correct option: B

A square

The correct answer is a square because all sides are equal and the diagonals are equal, indicating that PQRS is a square.

475 Easy

475 of 612

Application +4 | -1

S = \left\{ {z \in C:\arg \left( {\frac{{z - 2}}{{z + 2}}} \right) = \frac{\pi }{3}} \right\}

, then

S

View answer and explanation

Correct option: C

A circle

The correct answer is C (A circle) because the given condition represents a circle in the complex plane.

476 Easy

476 of 612

Application +4 | -1

The smallest positive integer

n

for which

\left(\frac{1+i}{1-i}\right)^n=-1

View answer and explanation

Correct option: B

Option B (2) is correct because when n=2, ((1+i)/(1-i))^n simplifies to -1, satisfying the given condition.

477 Easy

477 of 612

Application +4 | -1

z_k=\cos \left(\frac{k \pi}{10}\right)+i \sin \left(\frac{k \pi}{10}\right)

, then

z_1 z_2 z_3 z_4

is equal to

View answer and explanation

Correct option: A

-1

Option A (-1) is correct because by De Moivre's theorem, the product simplifies to

\cos(\pi) + i \sin(\pi)

, which equals -1.

478 Easy

478 of 612

Application +4 | -1

The value of

\sum\limits_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right)(

where

i=\sqrt{-1})

View answer and explanation

Correct option: D

-i

Option D (-i) is correct because the sum simplifies to -i using the properties of complex numbers and roots of unity.

479 Easy

479 of 612

Application +4 | -1

z=x+i y

is such that

|z-4|<|z-2|

, then

View answer and explanation

Correct option: D

x>3

and

y

is any real number

Option D (

x>3

and

y

is any real number) is correct because simplifying the inequality

|z-4|<|z-2|

leads to the condition

x>3

, with no restrictions on

y

480 Easy

480 of 612

Application +4 | -1

Let

{\left( { - 2 - \frac{1}{3}i} \right)^3} = \frac{{x + iy}}{{27}}\left( {i = \sqrt { - 1} } \right),

where x and y are real numbers, then y – x equals

View answer and explanation

Correct option: A

Option A (91) is correct because after expanding the complex number and equating real and imaginary parts, y - x equals 91.

Revision notes and key takeaways

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

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