If ${\tan ^{ - 1}}\left( {{\rm{\alpha }} + i{\rm{\beta }}} \right) = x + iy$, then $x$ is equal to

$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{2{\rm{\alpha }}}}{{1 - {{\rm{\alpha }}^2} - {{\rm{\beta }}^2}}}} \right)$

$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{2{\rm{\alpha }}}}{{1 + {{\rm{\alpha }}^2} + {{\rm{\beta }}^2}}}} \right)$

${\tan ^{ - 1}}\left( {\frac{{2{\rm{\alpha }}}}{{1 - {{\rm{\alpha }}^2} - {{\rm{\beta }}^2}}}} \right)$

If $|z|=1$ and $z \neq 0,$ then the points $z+\frac{1}{z}$ lie on a

Straight line not through the origin

Line segment lying on the y-axis

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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 61-90 of 337 questions

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61 Medium

61 of 337

Application +4 | -1

Let complex number

z=(a+\cos \theta)+(2 a-\sin \theta) i

. If

|z| \leq 2

for any

\theta \in R

, then the range of

a

View answer and explanation

Correct option: B

\left[-\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right]

Option B (

\left[-\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right]

) is correct because it represents the range of 'a' for which the magnitude of the complex number is less than or equal to 2 for all values of

\theta

62 Medium

62 of 337

Application +4 | -1

A\left(z_1\right), B\left(z_2\right)

and

C\left(z_3\right)

are the vertices of a triangle such that

\left|z_1\right|=\left|z_2\right|=\left|z_3\right|>0

, then orthocentre of

\triangle A B C

View answer and explanation

Correct option: A

z_1+z_2+z_3

Option A (

z_1+z_2+z_3

) is correct because the orthocenter is related to the centroid and circumcenter by the formula z_H = 3z_G - 2z_S, and since the circumcenter is 0, z_H = 3 * (1/3)(z_1 + z_2 + z_3).

63 Medium

63 of 337

Application +4 | -1

\frac{3}{2+\cos \theta+i \sin \theta}=a+i b

, then

(a-2)^2+b^2

is equal to

View answer and explanation

Correct option: B

The value of

(a-2)^2 + b^2

is 1 because it is obtained by manipulating the given expression and using trigonometric identities.

64 Medium

64 of 337

Application +4 | -1

In the argand plane, the complex number

z = 4 - 3i

is turned in the clockwise sense through

{180^0}

and stretched three times. The complex number represented by the new number is

View answer and explanation

Correct option: D

- 12 + 9i

Option D (

- 12 + 9i

) is correct because the complex number is rotated 180 degrees clockwise and then stretched by a factor of 3.

65 Medium

65 of 337

Application +4 | -1

x+i y=\frac{u+i v}{u-i v}

, then

x^2+y^2

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because multiplying

x+iy

by its conjugate

x-iy

results in 1.

66 Medium

66 of 337

Analyzing +4 | -1

Let

u = \frac{{2z + i}}{{z - ki}},\,\,z = x + iy\mathrm{ }\,\,\mathrm{and }\,\,k > 0.

if the curve represented by

{\mathop\mathrm{Re}\nolimits} (u) + {\mathop\mathrm{Im}\nolimits} (u) = 1

intersects the Y-axis at the point

P\mathrm{ }\,\mathrm{and}\,\mathrm{ Q}

where

PQ=5

then the value of k is

View answer and explanation

Correct option: C

Option C (2) is correct because solving the equation derived from the given conditions yields k=2, considering k>0.

67 Medium

67 of 337

Application +4 | -1

Let

z

be a complex number such that

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

. Then

z

lies on the circle of radius 2 and centre

View answer and explanation

Correct option: D

(0,-2)

Option D ((0,-2)) is correct because the equation represents a circle with center (0, -2) after simplification.

68 Medium

68 of 337

Application +4 | -1

Let

\mathrm{z}

and

\mathrm{w}

be non-zero complex numbers such that

|\mathrm{z}|=|\mathrm{w}|

and

\arg (\mathrm{z})+\arg (\mathrm{w})=\pi

. Then

\mathrm{z}=

View answer and explanation

Correct option: D

-\overline{\mathrm{w}}

Option D (

-\overline{\mathrm{w}}

) is correct because given the conditions, z is the negative of the conjugate of w.

69 Medium

69 of 337

Application +4 | -1

If the imaginary part of

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

-2 ,

then the locus of the point representing

z

in the complex plane is

View answer and explanation

Correct option: B

A straight line

Option B (A straight line) is correct because after substituting z = x + iy and simplifying, the imaginary part condition leads to a linear equation in x and y.

70 Medium

70 of 337

Analyzing +4 | -1

The number of integral solutions of

\frac{{x + 1}}{{{x^2} + 2}} > \frac{1}{4},

View answer and explanation

Correct option: C

Option C (5) is correct because solving the inequality

\frac{{x + 1}}{{{x^2} + 2}} > \frac{1}{4}

leads to the integral solutions 2, 3, 4, 5, and 6.

71 Medium

71 of 337

Analyzing +4 | -1

a,\;b,\;c \ne 0

and belongs to the set

\left\{ {0,\;1,2,\;3,\; \ldots \ldots ..,\;9} \right\}

,
Then

{\log _{10}}\left( {\frac{{a + 10b + {{10}^2}c}}{{{{10}^{ - 4}}a + {{10}^{ - 3}}b + {{10}^{ - 2}}c}}} \right)

is equal to

View answer and explanation

Correct option: D

Option D (4) is correct because the expression simplifies to

{\log _{10}}{10^4}

, which equals 4.

72 Medium

72 of 337

Application +4 | -1

\tan ^{-1}(\alpha+i \beta)=x+i y

, then

x

is equal to

View answer and explanation

Correct option: B

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

Option A (

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

) is correct because using the tangent addition formula and taking the real part leads to the correct expression for

x

73 Medium

73 of 337

Application +4 | -1

\left|z-\frac{4}{z}\right|=2

, then find the maximum value of

|z|

is equal to

View answer and explanation

Correct option: D

\sqrt{5}+1

The correct answer is D (

\sqrt{5}+1

) because it represents the maximum value of |z| derived from the given inequality using the triangle inequality.

74 Medium

74 of 337

Application +4 | -1

{\tan ^{ - 1}}\left( {\mathrm{\alpha } + i\mathrm{\beta }} \right) = x + iy

, then

x

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{2\mathrm{\alpha }}}{{1 - {\mathrm{\alpha }^2} - {\mathrm{\beta }^2}}}} \right)

Option A

\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{2\mathrm{\alpha }}}{{\left( {1 - {\mathrm{\alpha }^2} - {\mathrm{\beta }^2}} \right)}}} \right)

is the correct expression for x, derived by using the tangent addition formula and properties of complex conjugates.

75 Medium

75 of 337

Analyzing +4 | -1

z

is any complex number satisfying

|z-1|=1

then which of the following is correct?

View answer and explanation

Correct option: A

\arg (z-1)=2 \arg (z)

Option A (

\arg (z-1)=2 \arg (z)

) is correct because, by letting

z-1=e^{i \theta}

, we can express

z

in terms of

\theta

and show that

\arg(z) = \frac{{\theta}}{{2}}

, which leads to

\arg(z-1) = 2 \arg(z)

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76 Medium

76 of 337

Application +4 | -1

Z_1=\frac{1}{a+i}, a \neq 0

and

Z_2=\frac{1}{1+\mathrm{bi}} \quad b \neq 0

such that

Z_1=\bar{Z}_2

then

View answer and explanation

Correct option: C

\mathrm{a}=1, \mathrm{~b}=-1

Option C (a=1, b=-1) is correct because it satisfies the condition Z1 = conjugate(Z2) by equating real and imaginary parts.

77 Medium

77 of 337

Application +4 | -1

|z|=1

and

z \neq 0,

then the points

z+\frac{1}{z}

lie on a

View answer and explanation

Correct option: C

Line segment lying on the x-axis

Option C is correct. Since |z|=1, z can be represented as e^(iθ). Therefore, z + 1/z = 2cos(θ), which is a real number between -2 and 2, representing a line segment on the x-axis.

78 Medium

78 of 337

Application +4 | -1

Let

a=\operatorname{Im}\left(\frac{1+z^{2}}{2 i z}\right)

, where

z

is any non-zero complex number.

The set

\mathrm{A}=\{a:|z|=1

and

z \neq \pm 1\}

is equal to:

View answer and explanation

Correct option: A

(-1,1)

Option A ((-1,1)) is correct because by substituting z = x + iy and |z| = 1 into the expression for 'a', we find that a = x, and since z ≠ ±1, x lies in the open interval (-1, 1).

79 Medium

79 of 337

Application +4 | -1

If the cube roots of unity are

1, \omega, \omega^{2}

then the roots of the equation

(x-1)^{3}+8=0

, are

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it represents the roots of the given equation after simplifying and using the properties of cube roots of unity.

80 Medium

80 of 337

Application +4 | -1

\left( {{a_1} + i\;{b_1}} \right)\left( {{a_2} + i\;{b_2}} \right) \ldots \left( {{a_n} + i\;{b_n}} \right) = A + i\;B,

then

\left( {a_1^2 + b_1^2} \right)\left( {a_2^2 + b_2^2} \right) \ldots \left( {a_n^2 + b_n^2} \right)

is equal to

View answer and explanation

Correct option: B

{A^2} + {B^2}

Option B (

{A^2} + {B^2}

) is correct because the magnitude of the product of complex numbers is the product of their magnitudes.

81 Medium

81 of 337

Application +4 | -1

z

satisfies the equation

|z|-z=1+2 i

, then

z

is equal to

View answer and explanation

Correct option: B

\frac{3}{2}-2 i

Option B (

\frac{3}{2}-2 i

) is correct because it satisfies the given equation

|z|-z=1+2i

when substituting

z=x+iy

82 Medium

82 of 337

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.

83 Medium

83 of 337

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.

84 Medium

84 of 337

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)

85 Medium

85 of 337

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.

86 Medium

86 of 337

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.

87 Medium

87 of 337

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.

88 Medium

88 of 337

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|.

89 Medium

89 of 337

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.

90 Medium

90 of 337

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.

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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