Let ${z_1}$ and ${z_2}$ be two complex numbers with $\alpha $ and $\beta $ as their principle arguments such that ${\rm{\alpha }} + {\rm{\beta }} > {\rm{\pi }},$ then principal $\arg \left( {{z_1}{z_2}} \right)\;$ is given by

${\rm{\alpha }} + {\rm{\beta }} - {\rm{\pi }}$

${\rm{\alpha }} + {\rm{\beta }} + {\rm{\pi }}$

${\rm{\alpha }} + {\rm{\beta }} + 2{\rm{\pi }}$

${\rm{\alpha }} + {\rm{\beta }}$

Let a circle $C$ in complex plane pass through the points ${z_1} = 3 + 4i,\,\,{z_2} = 4 + 3i$ and $\,{z_3} = 5i.\,$ If $z( \ne {z_1})$ sis point on $C$ such that the line through $z$ and ${z_1}$ is perpendicular to the line through ${z_2}$ and ${z_3}$ then $\arg (z)$ is equal to:

${\tan ^{ - 1}}\left( {\frac{2}{{\sqrt 5 }}} \right) - \pi $

${\tan ^{ - 1}}\left( {\frac{3}{4}} \right) - \pi $

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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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211 Medium

211 of 979

Application +4 | -1

m

and

x

are two real numbers, then

e^{2 m i \cot ^{-1} x}\left(\frac{x i+1}{x i-1}\right)^m

(where

i=\sqrt{-1}

) is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because the expression simplifies to 1 using trigonometric identities and Euler's formula.

212 Easy

212 of 979

Application +4 | -1

\omega=e^{-2 \pi i / 3}, a, b, c, x, y, z

are non-zero complex numbers such that

a+b+c=x, a+b \omega+c \omega^2=y

and

a+

b \omega^2+c \omega=z

, then

\frac{|x|^2+|y|^2+|z|^2}{|a|^2+|b|^2+||^2}=

View answer and explanation

Correct option: D

Option D (9) is correct because, given the conditions and substituting a=b=c, x=3a, y=0, z=0, the expression simplifies to 9.

213 Easy

213 of 979

Application +4 | -1

{7^{{{\log }_7}\left( {{x^2} - 4\;x + 5} \right)}} = x - 1,x

may have values

View answer and explanation

Correct option: A

2,\;3

Option A (2, 3) is correct because these values satisfy the given equation when substituted and checked for validity within the logarithmic function.

214 Medium

214 of 979

Application +4 | -1

Let

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

. Then value of

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

\theta=3^{\circ}

View answer and explanation

Correct option: C

\frac{1}{2 \sin 3^{\circ}}

Option C (

\frac{1}{2 \sin 3^{\circ}}

) is correct because the summation simplifies to an imaginary part involving a geometric series, which can be evaluated using the formula for the sum of a geometric series.

215 Easy

215 of 979

Application +4 | -1

The value of

(1+i)^5(1-i)^5

2^n,

n

' is equal to

View answer and explanation

Correct option: A

Option A (5) is correct because

(1+i)^5(1-i)^5

simplifies to

2^5

, thus n=5.

216 Easy

216 of 979

Application +4 | -1

The value of

\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1

View answer and explanation

Correct option: B

-2

Option C (-2) is correct because after simplification, the expression equals -2.

217 Easy

217 of 979

Application +4 | -1

z_1=2+3 i

and

z_2=3-2 i

, then

z_1-z_2

equals to

-1+ bi.

The value of '

b

' is

View answer and explanation

Correct option: A

Option A (5) is correct because subtracting z2 from z1 results in -1 + 5i, so the value of 'b' is 5.

218 Medium

218 of 979

Application +4 | -1

The square roots of

-2+2 \sqrt{3} i

are

View answer and explanation

Correct option: A

\pm(1+\sqrt{3} i)

Option A is correct because the square roots of

-2+2\sqrt{3}i

are

\pm(1+\sqrt{3}i)

219 Easy

219 of 979

Understanding +4 | -1

z \neq 0

, then

\int\limits_{{x=0}}^{100}[\arg |z|] d x

is (where

[*]

denotes the greatest integer function)

View answer and explanation

Correct option: A

Option A is correct because the argument of the absolute value of any non-zero complex number is always 0, and the greatest integer function of 0 is 0.

220 Easy

220 of 979

Application +4 | -1

The mirror image of the curve

\arg \left(\frac{z+i}{z-1}\right)=\frac{\pi}{4}

i=\sqrt{-1}

in the line

x-y=0

, is

View answer and explanation

Correct option: C

\arg \left(\frac{z-i}{z+1}\right)=\frac{\pi}{4}

Option C (

\arg \left(\frac{{z - i}}{{z + 1}}\right) = \frac{\pi }{4}

) is correct because the mirror image of a complex number z in the line x-y=0 is obtained by swapping the real and imaginary parts, which is equivalent to taking

i\bar{z}

221 Easy

221 of 979

Application +4 | -1

Let

z

be a complex number such that

\left| {\frac{{z - i}}{{z + 2i}}} \right| = 1

and

\left| z \right| = \frac{5}{2}.

Then the value of

\left| {z + 3i} \right|

View answer and explanation

Correct option: B

\frac{7}{2}

Option B (7/2) is correct because, after finding the locus of z and using the given modulus condition, we can determine |z + 3i| = 7/2.

222 Easy

222 of 979

Application +4 | -1

|z|=z+3-2 i

, then

z

equals

View answer and explanation

Correct option: C

-5 / 6+2 i

Option C (-5/6 + 2i) is correct because solving the equation |z| = z + 3 - 2i for z yields z = -5/6 + 2i.

223 Medium

223 of 979

Application +4 | -1

\operatorname{Im}\left(\frac{z+2 i}{z+2}\right)=0

, then

z

lies on the curve

View answer and explanation

Correct option: C

x+y+2=0

Option C (

x+y+2=0

) is correct because the imaginary part of the given expression being zero implies that z lies on the line x+y+2=0.

224 Easy

224 of 979

Application +4 | -1

{z_1} = \sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)

And

{z_2} = \sqrt 3 \left( {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right)

, then

\left| {{z_1}{z_2}} \right|

View answer and explanation

Correct option: C

\sqrt 6

The correct answer is C (

\sqrt 6

) because the modulus of the product of two complex numbers is the product of their moduli.

225 Medium

225 of 979

Application +4 | -1

3^{49}(x+i y)=\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{100}

and

x=k y

, then

k

View answer and explanation

Correct option: D

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

The value of k is

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

because it is found by simplifying the given complex number equation and comparing the real and imaginary parts.

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226 Easy

226 of 979

Analyzing +4 | -1

Let

z\left( { \ne 2} \right)

be a complex number such that

{\log _{1/2}}\left| {z - 2} \right| > {\log _{1/2}}\left| z \right|,

then

View answer and explanation

Correct option: A

\mathrm{Re\;}\left( z \right) > 1

Option A (

\mathrm{Re\;}\left( z \right) > 1

) is correct because the inequality implies that the distance from z to 2 is less than the distance from z to 0, meaning z lies to the right of the perpendicular bisector of the segment joining 0 and 2.

227 Easy

227 of 979

Application +4 | -1

The conjugate of complex number

\frac{{2 - 3i}}{{4 - i}}

View answer and explanation

Correct option: B

\frac{{11 + 10i}}{{17}}

Option B (

\frac{{11 + 10i}}{{17}}

) is correct because it is the result of finding the conjugate of the given complex number and simplifying the expression.

228 Medium

228 of 979

Application +4 | -1

x_r=\cos \left(\pi / 3^r\right)

- i

\sin \left(\pi / 3^r\right)

, (where

i=\sqrt{-1}

). then value of

x_1 \cdot x_2 \cdot \ldots \infty

, is

View answer and explanation

Correct option: C

-{i}

Option C (-i) is correct because the product simplifies to -i using the properties of complex numbers and infinite geometric series.

229 Easy

229 of 979

Application +4 | -1

\omega

is a complex cube root of unity, then

225 + {\left( {3\omega + 8{\omega ^2}} \right)^2} + {\left( {3{\omega ^2} + 8\omega } \right)^2}

is equal to

View answer and explanation

Correct option: D

248

Option D (248) is correct because it is the result of simplifying the expression using the properties of complex cube roots of unity.

230 Easy

230 of 979

Application +4 | -1

Let

z

be a complex number such that

\left| z \right| + z = 3 + i

(where

i = \sqrt { - 1} )

. Then

\left| z \right|

is equal to

View answer and explanation

Correct option: B

\frac{5}{3}

Option B (5/3) is correct because by substituting

z = x + i

into the given equation and solving for x, we can find the magnitude of z.

231 Easy

231 of 979

Understanding +4 | -1

Let

{z_1}

and

{z_2}

be two complex numbers with

\alpha

and

\beta

as their principle arguments such that

\mathrm{\alpha } + \mathrm{\beta } > \mathrm{\pi },

then principal

\arg \left( {{z_1}{z_2}} \right)\;

is given by

View answer and explanation

Correct option: B

\mathrm{\alpha } + \mathrm{\beta } - \mathrm{\pi }

Option B (

\alpha + \beta - \pi

) is correct because the principal argument must lie between

-\pi

and

\pi

, and since

\alpha + \beta > \pi

, we subtract

2\pi

to bring it into the range.

232 Easy

232 of 979

Understanding +4 | -1

The complex numbers

{z_1},{z_2},{z_3},{z_4}

taken in that order in the Argand plane represent the vertices of a parallelogram iff

View answer and explanation

Correct option: B

{z_1} + {z_3} = {z_2} + {z_4}

Option B (

{z_1} + {z_3} = {z_2} + {z_4}

) is correct because in a parallelogram, the sum of opposite vertices are equal.

233 Easy

233 of 979

Application +4 | -1

1,\;\omega ,\;{\omega ^2}

are the cube roots of unity, then

{\left( {3 + {\omega ^2} + {\omega ^4}} \right)^6}

is equal to

View answer and explanation

Correct option: A

Option A (64) is correct because simplifying the expression using the properties of cube roots of unity results in 2⁶ = 64.

234 Hard

234 of 979

Analyzing +4 | -1

The value of the expression

2\left( {1 + \frac{1}{\omega }} \right)\left( {1 + \frac{1}{{{\omega ^2}}}} \right) + 3\left( {2 + \frac{1}{\omega }} \right)\left( {2 + \frac{1}{{{\omega ^2}}}} \right) + \ldots + \left( {n + 1} \right)\left( {n + \frac{1}{\omega }} \right)\left( {n + \frac{1}{{{\omega ^2}}}} \right)

View answer and explanation

Correct option: C

{\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} + n

Option C (

{\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} + n

) is correct because by simplifying the expression using properties of complex cube roots of unity, we can derive the given result.

235 Easy

235 of 979

Analyzing +4 | -1

Let a circle

C

in complex plane pass through the points

{z_1} = 3 + 4i,\,\,{z_2} = 4 + 3i

and

\,{z_3} = 5i.\,

z( \ne {z_1})

sis point on

C

such that the line through

z

and

{z_1}

is perpendicular to the line through

{z_2}

and

{z_3}

then

\arg (z)

is equal to:

View answer and explanation

Correct option: B

{\tan ^{ - 1}}\left( {\frac{{24}}{7}} \right) - \pi

The correct answer is B (

{\tan ^{ - 1}}\left( {\frac{{24}}{7}} \right) - \pi

) because it is the argument of the point z on the circle, calculated using the given conditions and complex number properties.

236 Easy

236 of 979

Application +4 | -1

\omega

and

{\omega ^2}

are the two imaginary cube roots of unity, then the equation whose roots are

a{\omega ^{317}}

and

a{\omega ^{382}}

, is

View answer and explanation

Correct option: A

{x^2} + ax + {a^2} = 0

Option A (

{x^2} + ax + {a^2} = 0

) is correct because it is the quadratic equation formed using the given roots and properties of cube roots of unity.

237 Medium

237 of 979

Application +4 | -1

z \neq 0

be a complex number such that

\left|z-\frac{1}{z}\right|=2

, then the maximum value of

|z|

View answer and explanation

Correct option: D

\sqrt{2}+1

Option D (

\sqrt{{2}}+1

) is correct because by using the triangle inequality and solving the resulting quadratic, we find the maximum value of |z|.

238 Medium

238 of 979

Application +4 | -1

z_1=\sqrt{3}+i \sqrt{3}

and

z_2=\sqrt{3}+i

, then the complex number

\left(\frac{z_1}{z_2}\right)^{50}

lies in the quadrant number

View answer and explanation

Correct option: A

Option A (I) is correct because after simplification, the complex number lies in the first quadrant.

239 Medium

239 of 979

Analyzing +4 | -1

The locus of the complex number

z

in an argand plane satisfying the inequality

\log _{1 / 2}\left(\frac{|z-1|+4}{3|z-1|-2}\right) > 1\left(\right.

where,

\left.|z-1| \neq \frac{2}{3}\right)

View answer and explanation

Correct option: C

exterior of a circle

Option C (exterior of a circle) is correct because the inequality simplifies to

|z-1| > 3

, which represents the region outside a circle centered at 1 with radius 3.

240 Easy

240 of 979

Understanding +4 | -1

arg

\left( {\bar z} \right)

is equal to

View answer and explanation

Correct option: B

2\mathrm{\pi } - \arg \left( z \right)

Option B (2π - arg(z)) is correct because the argument of the conjugate of a complex number is the negative of the argument of the original number, or 2π minus the argument.

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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