If $a = \sqrt {2\;i} ,$ then which of the following is correct?

$a = - \left( {\sqrt 2 } \right)i$

If $\cos \alpha + 2\cos {\rm{\beta }} + 3\cos {\rm{\gamma }} = \sin \alpha + 2\sin {\rm{\beta }} + 3\sin {\rm{\gamma }} = 0,$ then the value of $\sin 3\;\alpha + 8\sin 3\;{\rm{\beta }} + 27\sin 3\;{\rm{\gamma }}$ is

$\sin \left( {{\rm{\alpha }} + {\rm{\beta }} + {\rm{\gamma }}} \right)$

$3\sin \left( {{\rm{\alpha }} + {\rm{\beta }} + {\rm{\gamma }}} \right)$

$\sin \left( {{\rm{\alpha }} + 2{\rm{\beta }} + 3{\rm{\gamma }}} \right)$

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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Mixed difficulty

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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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331 Easy

331 of 979

Application +4 | -1

z \in C

, then minimum value of

|z-2+3 i|+

|z-1+i|

View answer and explanation

Correct option: A

\sqrt{5}

Option A (

\sqrt{5}

) is correct because the minimum value is found using the triangle inequality property of complex numbers.

332 Medium

332 of 979

Application +4 | -1

z_1 \& z_2

are two complex numbers satisfying

|z-4|=\operatorname{Re}(z)

and having greatest and least argument respectively, then area of triangle formed by origin,

z_1 \& z_2

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,

z_1

, and

z_2

is calculated using the coordinates of

z_1

and

z_2

which are derived from the given condition.

333 Easy

333 of 979

Application +4 | -1

Suppose

a < 0

and

z_1, z_2, z_3, z_4

be the fourth roots of

a

. Then

z_1^2+z_2^2+z_3^2+z_4^2

is equal

View answer and explanation

Correct option: C

a+|a|

Option C (a+|a|) is correct because the sum of the squares of the fourth roots of a negative number is 0, which is equivalent to a + |a| when a < 0.

334 Easy

334 of 979

Application +4 | -1

A real value of

x

will satisfy the equation

\left( {\frac{{3 - 4ix}}{{3 + 4ix}}} \right) = \alpha - i\beta \;\left( {\alpha ,\beta \;\mathrm{are\;real}} \right)

View answer and explanation

Correct option: C

{\alpha ^2} + {\beta ^2} = 1

Option C (

{\alpha ^2} + {\beta ^2} = 1

) is correct because by taking the modulus of both sides of the given equation and simplifying, we find that

{\alpha ^2} + {\beta ^2} = 1

335 Easy

335 of 979

Application +4 | -1

a = \sqrt {2\;i} ,

then which of the following is correct?

View answer and explanation

Correct option: A

a = 1 + i

Option A (

a = 1 + i

) is correct because solving

\sqrt{2i}

yields

\pm(1+i)

, and

a = 1 + i

is one of the possible solutions.

336 Easy

336 of 979

Application +4 | -1

z = \frac{4}{{1 - i}}

, then

\bar z

is (where

\bar z

is complex conjugate of

z

)

View answer and explanation

Correct option: D

\frac{4}{{1 + i}}

Option D

\frac{4}{{1 + i}}

is correct because the conjugate of a complex number involves changing the sign of the imaginary part.

337 Easy

337 of 979

Application +4 | -1

For

x_1, x_2, y_1, y_2, \in R

, if

0 < x_1, < x_2, y_1=y_2

and

z_1=x_1+i y_1, z_2=x_2+i y_2

and

z_3=\frac{1}{2}\left(z_1+z_2\right)

then

z_1, z_2, z_3

satisfy

View answer and explanation

Correct option: D

\left|z_1\right| < \left|z_3\right| < \left|z_2\right|

Option D (

\left|z_1\right| < \left|z_3\right| < \left|z_2\right|

) is correct because

z_3

is the midpoint of

z_1

and

z_2

, and

x_1 < x_2

which implies

|z_1| < |z_3| < |z_2|

338 Medium

338 of 979

Application +4 | -1

The range of real number

\alpha

for which the equation

z+\alpha|z-1|+2 i=0

has a solution, is

View answer and explanation

Correct option: A

\left[-\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{2}\right]

Option A

\left[-\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{2}\right]

is correct because it represents the range of

\alpha

for which the quadratic equation has real solutions.

339 Easy

339 of 979

Application +4 | -1

The curve represented by

\mathrm{Im}\left( {{z^2}} \right) = k,

where

k

is a non-zero real number, is

View answer and explanation

Correct option: D

A hyperbola

Option D (A hyperbola) is correct because the equation Im(

{z^2}

) = k, where k is a non-zero real number, simplifies to xy = k/2, which represents a hyperbola.

340 Easy

340 of 979

Application +4 | -1

z=2+3 i

, then

z^5+(\bar{z})^5

is equal to :

View answer and explanation

Correct option: A

244

Option A (244) is correct because expanding z⁵ and (z_conjugate)⁵ and adding them results in 244.

341 Easy

341 of 979

Understanding +4 | -1

z

be complex number such that equation

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

always represents an ellipse, then range of a

\left(\in R^{+}\right)

View answer and explanation

Correct option: D

(0,3)

Option D (

(0,3)

) is correct because the given condition represents an ellipse only if the distance between the foci is less than the length of the major axis.

342 Easy

342 of 979

Application +4 | -1

If square root of

- 7 + 24i

x + iy

, then

x

View answer and explanation

Correct option: C

\pm 3

Option C (

\pm 3

) is correct because it represents the real part of the square root of the complex number -7 + 24i, calculated using the appropriate formula.

343 Easy

343 of 979

Application +4 | -1

Let

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

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

\sum \limits_{m = 1}^{15} \mathrm{Im}\left( {{z^{2m - 1}}} \right)\,\mathrm{at}\,\,\theta = {2^0}

View answer and explanation

Correct option: D

\frac{1}{{4\sin {2^0}}}

Option D (

\frac{1}{{4\sin {2^0}}}

) is correct because the summation of the imaginary parts of the complex number series simplifies to this expression using trigonometric identities.

344 Medium

344 of 979

Application +4 | -1

If complex number

z

satisfies

|z-6 i|=\operatorname{Im}(z)

, then range of

(\arg z-\arg \bar{z})

will be

View answer and explanation

Correct option: A

\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]

Option A ([

\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.

345 Medium

345 of 979

Application +4 | -1

Let

a

be a complex number such that

\left| a \right| < 1

and

{z_1},\;{z_2}, \ldots

be vertices of a polygon such that

{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

\left| {z - \frac{1}{{1 - a}}} \right| = \frac{1}{{\left| {1 - a} \right|}}

Option C (

\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|.

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346 Medium

346 of 979

Analyzing +4 | -1

Consider an ellipse having its foci at

A\left(z_1\right)

and

B\left(z_2\right)

in the argand plane. If the eccentricity of the ellipse is

e

and it is known that origin is an interior point of the ellipse, then

View answer and explanation

Correct option: B

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

e

must satisfy the condition

e \in (0, \frac{{|z_1-z_2|}}{{|z_1|+|z_2|}})

for the origin to be an interior point of the ellipse.

347 Easy

347 of 979

Application +4 | -1

The values of

x

satisfying

\left| {x - 4} \right| + \left| {x - 9} \right| = 5

View answer and explanation

Correct option: A

x = 4,\;9

The correct answer is A (

x = 4,\;9

) because these are the only values that satisfy the given absolute value equation.

348 Easy

348 of 979

Application +4 | -1

\omega(\neq 1)

is a cube root of unity and

(1+\omega)^{2017}=

A+B \omega

. Then

A

and

B

are respectively the numbers

View answer and explanation

Correct option: B

1,1

Option B (1,1) is correct because

(1+\omega)^{2017}

simplifies to

-\omega^2

, which can be rewritten as

1 + \omega

, thus A=1 and B=1.

349 Easy

349 of 979

Analyzing +4 | -1

\cos \alpha + 2\cos \mathrm{\beta } + 3\cos \mathrm{\gamma } = \sin \alpha + 2\sin \mathrm{\beta } + 3\sin \mathrm{\gamma } = 0,

then the value of

\sin 3\;\alpha + 8\sin 3\;\mathrm{\beta } + 27\sin 3\;\mathrm{\gamma }

View answer and explanation

Correct option: C

18\sin \left( {\alpha + \beta + \gamma } \right)

Option C (

18\sin \left( {\alpha + \beta + \gamma } \right)

) is correct because it is derived from complex number manipulation and trigonometric identities given the initial conditions.

350 Medium

350 of 979

Application +4 | -1

The curve in the complex plane given by the equation

\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)

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.

351 Easy

351 of 979

Application +4 | -1

Let

z_1, z_2, z_3

be three complex number such that

z_1+z_2+z_3=0

and

\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1

, then

\left|z_1^2+2 z_2^2+z_3^2\right|

equals

View answer and explanation

Correct option: A

Option A (1) is correct because using the given conditions, we can simplify the expression to |z_2|^2 which is equal to 1.

352 Medium

352 of 979

Understanding +4 | -1

The equation

z\bar z + a\bar z + \bar az + b = 0,\;\;b \in R

represents a circle, if

View answer and explanation

Correct option: 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).

353 Medium

353 of 979

Application +4 | -1

How many solutions are possible for

\operatorname{Re}(z)^2=0

and

|z|=a \sqrt{2}(a>0)?

View answer and explanation

Correct option: D

Option D (4) is correct because there are four distinct solutions that satisfy both the real part condition and the magnitude condition.

354 Easy

354 of 979

Application +4 | -1

{\log _{\sqrt 3 }}\left( {\frac{{{{\left| z \right|}^2} - \left| z \right| + 1}}{{2 + \left| z \right|}}} \right)

< 2

, then the locus of

z

View answer and explanation

Correct option: B

\left| z \right| < 5

Option B (

\left| z \right| < 5

) is correct because solving the inequality results in the condition that the modulus of

z

must be less than 5.

355 Medium

355 of 979

Application +4 | -1

if locus of point

p(z)

which satisfies

\arg (3+z+z+|z| i)=\frac{\pi}{4}

is a part of conic

\mathrm{C}

then length of latus rectumof

\mathrm{C}

View answer and explanation

Correct option: D

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.

356 Medium

356 of 979

Analyzing +4 | -1

\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

{x^2} - 5x + 4 = 0

Option C (

{x^2} - 5x + 4 = 0

) is correct because the sum and product of the given roots match the coefficients of this quadratic equation.

357 Medium

357 of 979

Application +4 | -1

The value of

\lambda

, for which the equation

6 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

\lambda

that makes the equation represent a pair of straight lines is

\frac{{15}}{{361}}

, which is not among the given options.

358 Medium

358 of 979

Application +4 | -1

\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.

359 Medium

359 of 979

Application +4 | -1

If 2 and 6 are the roots of the equation

ax^2+b x+1=0

, then the quadratic equation, whose roots are

\frac{1}{2 a+b}

and

\frac{1}{6 a+b}

, is:

View answer and explanation

Correct option: D

x^2+8 x+12=0

Option B (

\frac{\pi}{2}

) is correct because, after evaluating the integral using properties of definite integrals, the value of

\alpha

is found to be

\frac{\pi}{2}

360 Medium

360 of 979

Analyzing +4 | -1

Given that

a,\;b \in \left\{ {0,\;1,\;2, \ldots ,9} \right\}

with

a + b \ne 0

and that

{\left( {a + \frac{b}{{10}}} \right)^x} = {\left( {\frac{a}{{10}} + \frac{b}{{100}}} \right)^y} = 1000

. Then,

\frac{1}{x} - \frac{1}{y}

is equal to

View answer and explanation

Correct option: C

\frac{1}{3}

Option C (

\frac{1}{3}

) is correct because by setting a = 0 and b = 1, we can solve for x and y and find that

\frac{1}{x} - \frac{1}{y} = \frac{1}{3}

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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