If $z=a+i b$, where $a > 0, b > 0$, then

$|z| < \frac{1}{\sqrt{2}}(a+b)$

If $z = r\;{e^{i\;{\rm{\theta }}}},$ $\left| {{e^{i\;z}}} \right|$ is equal to

${e^{ - r\sin {\rm{\theta }}}}$

$r\;{e^{ - r\sin {\rm{\theta }}}}$

${e^{ - r\cos {\rm{\theta }}}}$

$r\;{e^{ - r\cos {\rm{\theta }}}}$

One of the square root of $6 + 4\sqrt 3 $ is

$\sqrt 3 \left( {\sqrt 3 + 1} \right)$

$ - \sqrt 3 \left( {\sqrt 3 - 1} \right)$

$\sqrt 3 \left( { - \sqrt 3 + 1} \right)$

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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Medium

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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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Showing 151-180 of 337 questions

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151 Medium

151 of 337

Application +4 | -1

x_1 \leq x_2

and

y_1 \leq y_2

for any two complex numbers

z_1=x_1+i y_1

and

z_2=x_2+i y_2

, then we say

z_1 \cap z_2.

Let

z

be a complex number with

1 \cap z

. Then

View answer and explanation

Correct option: C

\frac{1-z}{1+z} \cap 0

Option C (

\frac{{1-z}}{{1+z}} \cap 0

) is correct because given the condition 1 ∩ z, it follows that the real and imaginary parts of

\frac{{1-z}}{{1+z}}

are non-positive, implying it is \"less than or equal to\" 0.

152 Medium

152 of 337

Analyzing +4 | -1

p,q,r

are positive and are in AP, then roots of the quadratic equation

p{x^2} + qx + r = 0

are complex for

View answer and explanation

Correct option: B

\left| {\frac{p}{r} - 7} \right| < 4\sqrt 3

Option B is correct because the condition for complex roots of the quadratic equation, given that p, q, and r are in AP, leads to the inequality

\left| {\frac{{p}}{{r}} - 7} \right| < 4\sqrt 3

153 Medium

153 of 337

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: B

The expression

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

is equal to 1 because it simplifies to 1 using trigonometric identities and Euler's formula.

154 Medium

154 of 337

Analyzing +4 | -1

Let

z_1, z_2

and

z_3

are the points on the argand plane which lie on the circle with equation

\left|z-z_0\right|=\frac{4}{\sqrt{3}}

(where

z_0

is the centre of the circle). If

z_1=0, z_2=-4

and

z_3=4+3 z_0

, then

\left|\arg \left(z_0\right)\right|

is equal to (where

\arg Z \in(-\pi, \pi])

View answer and explanation

Correct option: C

\frac{5 \pi}{6}

Option C (

\frac{{5 \pi}}{{6}}

) is correct because by using the properties of equilateral triangles and complex numbers, we can find the argument of z_0.

155 Medium

155 of 337

Application +4 | -1

z=a+i b

, where

a > 0, b > 0

, then

View answer and explanation

Correct option: B

|z| \geq \frac{1}{\sqrt{2}}(a+b)

Option B is correct because, given

z=a+ib

where

a > 0, b > 0

, then

|z| \geq \frac{1}{\sqrt{2}}(a+b)

156 Medium

156 of 337

Analyzing +4 | -1

b+i c=(1+a) z

and

a^2+b^2+c^2=1

, then

\frac{1+i z}{1-i z}

equals

View answer and explanation

Correct option: D

\frac{a+i b}{1+c}

Option D (

\frac{{a+i b}}{{1+c}}

) is correct because, after substituting

z = \frac{{b+ic}}{{1+a}}

and simplifying, the expression

\frac{{1+iz}}{{1-iz}}

reduces to

\frac{{a+ib}}{{1+c}}

157 Medium

157 of 337

Application +4 | -1

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

, (where

\left.i=\sqrt{-1}\right)

. 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 infinite product simplifies to -i using trigonometric identities and limits.

158 Medium

158 of 337

Application +4 | -1

The relation between the real numbers a and

b

, which satisfy the equation

\frac{1-\mathrm{ix}}{1+\mathrm{ix}}=\mathrm{a}-\mathrm{ib}

, for some real value of

x

, is

View answer and explanation

Correct option: C

a^2+b^2=1

Option C (

a^2+b^2=1

) is correct because by taking the modulus of both sides of the equation and simplifying, we arrive at the relationship

a^2 + b^2 = 1

159 Medium

159 of 337

Application +4 | -1

Given

{z}

is a complex number such that

|z| \leq 2

, then the maximum value of

|i z+6-8 i|

is equal to-

View answer and explanation

Correct option: A

Option A (12) is correct because the maximum value of

|iz + 6 - 8i|

is found using the triangle inequality, resulting in

|iz| + |6-8i| = 2 + 10 = 12

160 Medium

160 of 337

Application +4 | -1

The magnitude and amplitude of

\frac{{\left( {1 + i\sqrt 3 } \right)\left( {2 + 2i} \right)}}{{\left( {\sqrt 3 - i} \right)}}

are respectively

View answer and explanation

Correct option: B

2\sqrt 2 ,\frac{{3\pi }}{4}

Option B (

2\sqrt 2 ,\frac{{3\pi }}{4}

) is correct because it represents the magnitude and amplitude of the given complex number after simplification.

161 Medium

161 of 337

Analyzing +4 | -1

z_1 \neq z_2

and

\left|z_1+z_2\right|=\left|\frac{1}{z_1}+\frac{1}{z_2}\right|

then

View answer and explanation

Correct option: B

z_1 \cdot z_2

is unimodular

Option B (

z_1 \cdot z_2

is unimodular) is correct because the given condition implies that

|z_1 z_2| = 1

, which means

z_1 z_2

is unimodular.

162 Medium

162 of 337

Understanding +4 | -1

\alpha, \beta

are two complex numbers, then

|\alpha|^2+|\beta|^2

is equal to

View answer and explanation

Correct option: B

\frac{1}{2}\left(|\alpha+\beta|^2+|\alpha-\beta|^2\right)

Option B (

\frac{1}{2}(|\alpha+\beta|^2+|\alpha-\beta|^2)

) is correct because expanding the magnitudes and simplifying leads to the given equality.

163 Medium

163 of 337

Application +4 | -1

\left(\frac{1+i \sin \frac{\pi}{8}+\cos \frac{\pi}{8}}{1-i \sin \frac{\pi}{8}+\cos \frac{\pi}{8}}\right)^8

is equal to

View answer and explanation

Correct option: C

-1

Option C (-1) is the correct answer because after simplification using trigonometric identities and De Moivre's theorem, the expression evaluates to -1.

164 Medium

164 of 337

Application +4 | -1

\arg (z) < 0

, then

\arg \left(\frac{z-\bar{z}}{2}\right)

is equal to

View answer and explanation

Correct option: C

-\frac{\pi}{2}

Option C (

-\frac{\pi}{2}

) is correct because if arg(z) < 0, then (z - conjugate(z))/2 is a purely imaginary number with a negative imaginary part, implying an argument of -

\frac{\pi}{2}

165 Medium

165 of 337

Application +4 | -1

z = r\;{e^{i\;\mathrm{\theta }}},

\left| {{e^{i\;z}}} \right|

is equal to

View answer and explanation

Correct option: A

{e^{ - r\sin \mathrm{\theta }}}

Option A (

{e^{ - r\sin \mathrm{\theta }}}

) is correct because it is derived from Euler's formula and properties of complex exponentials.

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166 Medium

166 of 337

Application +4 | -1

Dividing

f(z)

z-i

, we obtain the remainder

i

and dividing it by

z+i

, we get the remainder

1+i

. The remainder upon the division of

f(z)

z^2+1

, is

View answer and explanation

Correct option: B

\frac{1}{2}(i z+1)+i

Option B (

\frac{{1}}{{2}}(i z+1)+i

) is correct because it is the remainder obtained after dividing f(z) by

z^2 + 1

167 Medium

167 of 337

Application +4 | -1

Let

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

, where

\mathrm{x}

and

\mathrm{y}

are real numbers then

y-x

equals :

View answer and explanation

Correct option: A

Option A (91) is correct because expanding

\left(-2-\frac{1}{3} i\right)^{3}

, equating it to

\frac{x+i y}{27}

, and then calculating

y-x

yields 91.

168 Medium

168 of 337

Application +4 | -1

{\left( {x + iy} \right)^{1/3}} = 2 + 3i,

then

3x + 2y

is equal to

View answer and explanation

Correct option: C

- 120

Option C (-120) is correct because cubing (2+3i) gives x+iy = -46 + 9i, and then 3x + 2y = -120.

169 Medium

169 of 337

Application +4 | -1

The solution of the equation

2 z=|z|+2 i

, where

z

is a complex number, is

View answer and explanation

Correct option: B

{z}=\frac{\sqrt{3}}{3}+{i}

Option D (

{z}=\frac{\sqrt{{3}}}{{3}}+{{i}}

) is correct because it is the solution obtained by substituting

z = x + iy

into the given equation and solving for x and y.

170 Medium

170 of 337

Application +4 | -1

\frac{1-i x}{1+i x}=a-i b

and

a^2+b^2=1

, where

a

and

b

are real, then

x

is equal to

View answer and explanation

Correct option: B

\frac{2 b}{(1+a)^2+b^2}

Option B (

\frac{2 b}{(1+a)^2+b^2}

) is correct because rationalizing the denominator and equating real and imaginary parts allows us to express x in terms of a and b.

171 Medium

171 of 337

Analyzing +4 | -1

Z_n=\operatorname{cis}\left(\frac{\pi}{n(n+1)(n+2)}\right)

for

n=1,2,3 \ldots

. and

Z = \lim\limits_{n \to \infty } \left( {{z_1}{z_2} \ldots .{z_n}} \right)

, then principal argument of

Z

View answer and explanation

Correct option: C

\frac{\pi}{4}

Option C (

\frac{\pi}{4}

) is correct because the principal argument of the limit of the product of complex numbers

Z_n

\frac{\pi}{4}

172 Medium

172 of 337

Application +4 | -1

Let

z_1, z_2

are any complex numbers which satisfies the following conditions

\left|z_1\right|=12

and

\left|z_2-\sqrt{3}-i\right|=5

respectively then range of values of

\left|z_1-z_2\right|

will be

View answer and explanation

Correct option: B

[5,19]

Option B ([5,19]) is correct because by using the triangle inequality and the given conditions, we can find the range of values for

|z_1 - z_2|

173 Medium

173 of 337

Application +4 | -1

\mathrm{i}=\sqrt{-1}

, then

4+5\left(-\frac{1}{2}+\frac{\mathrm{i} \sqrt{3}}{2}\right)^{334}+3\left(-\frac{1}{2}+\frac{\mathrm{i} \sqrt{3}}{2}\right)^{365}

is equal to:

View answer and explanation

Correct option: C

i \sqrt{3}

Option C is correct because simplifying the expression using properties of complex numbers, specifically omega (

\omega

), leads to the result

i \sqrt{3}

174 Medium

174 of 337

Understanding +4 | -1

\left|Z_1\right|=\left|Z_2\right|

and

\arg \left(\frac{Z_1}{Z_2}\right)=\pi

then,

Z_1+Z_2=

View answer and explanation

Correct option: A

Option A (0) is correct because if the magnitudes are equal and the argument of their ratio is π, the complex numbers are additive inverses.

175 Medium

175 of 337

Application +4 | -1

One of the square root of

6 + 4\sqrt 3

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct. The square root of 6 + 4√3 cannot be expressed in the form √x + √y where x and y are rational numbers, as the discriminant becomes negative.

176 Medium

176 of 337

Analyzing +4 | -1

All the roots of the equation

a_1 z^3+a_2 z^2+a_3 z+a_4=3

, where

\left|a_j\right| \leq 1

i=1,2,3,4

, lie outside the circle with centre origin and radius

View answer and explanation

Correct option: C

\frac{1}{3}

Option C (

\frac{{1}}{{3}}

) is correct because all roots of the equation lie outside the circle with radius

\frac{{2}}{{3}}

, which means they lie outside the circle with radius

\frac{{1}}{{3}}

as well.

177 Medium

177 of 337

Application +4 | -1

z \neq 0

and

2+\cos \theta+i \sin \theta=3 / z

, then value of

2(z+\bar{z})-|z|^2

View answer and explanation

Correct option: C

Option C (3) is correct because by manipulating the given equation and using the properties of complex conjugates, we can isolate and solve for the desired expression.

178 Medium

178 of 337

Application +4 | -1

If for complex numbers

{z_1}

and

{z_2},\arg \left( {{z_1}} \right) - \arg \left( {{z_2}} \right) = 0,

then

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

| |{{z_1}} | - |{z_2}|| |

) is correct because when the arguments of two complex numbers are equal, the magnitude of their difference is the absolute difference of their magnitudes.

179 Medium

179 of 337

Application +4 | -1

Let

\alpha \mathrm{ }\,\mathrm{and}\,\mathrm{ }\beta

be two roots of the equation

{x^2} + 2x + 2 = 0,

then

{\alpha ^{15}} + {\beta ^{15}}

is equal to

View answer and explanation

Correct option: C

-256

Option C (-256) is correct because it is the value of the expression

{\alpha ^{15}} + {\beta ^{15}}

where

\alpha

and

\beta

are the roots of the given quadratic equation.

180 Medium

180 of 337

Application +4 | -1

For all complex numbers

z_1, z_2

satisfying

\left|z_1\right|=12

and

\left|z_2-3-4 i\right|=5

, the minimum value of

\left|z_1-z_2\right|

View answer and explanation

Correct option: B

Option B (2) is correct because the minimum value of |z1-z2| is found using the triangle inequality and given conditions.

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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