If $2 \cos \theta=x+\frac{1}{x}$ and $2 \cos \phi=y+\frac{1}{y}$, then

$\frac{x}{y}+\frac{y}{x}=2 \cos (\theta+\phi)$

$\frac{x^m}{y^n}+\frac{y^n}{x^m}=2 \cos (m \theta+n \phi)$

$x y+\frac{1}{x y}=2 \cos (\theta-\phi)$

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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Hard

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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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1 Hard

1 of 30

Application +4 | -1

1,\;{a_1},\;{a_2}, \ldots ,{a_{n - 1}}

are the

n

roots of unity, then the value of

\left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right)\left( {1 - {a_3}} \right) \ldots \left( {1 - {a_{n - 1}}} \right)

is equal to

View answer and explanation

Correct option: C

n

Option C (n) is correct because the expression is equal to the derivative of x^n - 1 evaluated at x=1, which equals n.

2 Hard

2 of 30

Application +4 | -1

The expression

\tan \left\{ {i\log \left( {\frac{{a - ib}}{{a + ib}}} \right)} \right\}

reduces to

View answer and explanation

Correct option: B

\frac{{2\;ab}}{{{a^2} - {b^2}}}

Option B (

\frac{{2\;ab}}{{{{a^2} - {b^2}}}}

) is correct because it represents the simplified form of the given expression after applying logarithmic and trigonometric identities.

3 Hard

3 of 30

Analyzing +4 | -1

{\log _{\tan 30^\circ }}\left( {\frac{{2{{\left| z \right|}^2} + 2\left| z \right| - 3}}{{\left| z \right| + 1}}} \right) < - 2,

then

View answer and explanation

Correct option: C

\left| z \right| > 2

Option C (

\left| z \right| > 2

) is correct because solving the logarithmic inequality leads to the condition that the magnitude of z is greater than 2.

4 Hard

4 of 30

Analyzing +4 | -1

\left| {z - 4 - 3i} \right| \le 1

and

m

and

n

are the least and greatest value of

\left| z \right|

and

\lambda

is the least value of

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

in the interval

\left( {0,\;\infty } \right)

then

\lambda

is equal to

View answer and explanation

Correct option: B

n

Option B (

n

) is correct because the least value of the given expression is 6, which is equal to the greatest value of |z|.

5 Hard

5 of 30

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.

6 Hard

6 of 30

Analyzing +4 | -1

If the expression

\frac{{\left[ {\sin \left( {\frac{x}{2}} \right) + \cos \left( {\frac{x}{2}} \right) - i\tan \left( x \right)} \right]}}{{\left[ {1 + 2i\sin \left( {\frac{x}{2}} \right)} \right]}}

is real then the set of all possible value of

x

View answer and explanation

Correct option: B

2n\pi

Option B (

2n\pi

) is the correct answer because it satisfies the condition for the given expression to be real by making the imaginary part zero.

7 Hard

7 of 30

Analyzing +4 | -1

z=x+i y

lies in III quadrant, then

\frac{\bar{z}}{z}

also lies in III quadrant, if

View answer and explanation

Correct option: C

y < x < 0

Option C (

y < x < 0

) is correct because for

\frac{{\bar{{z}}}}{{z}}

to lie in the third quadrant, both its real and imaginary parts must be negative, leading to the condition

y < x < 0

8 Hard

8 of 30

Analyzing +4 | -1

The common roots of the equations

{x^3} + 2{x^2} + 2x + 1 = 0

and

1 + {x^{2002}} + {x^{2003}} = 0

(where

\omega

is a complex cube root of unity)

View answer and explanation

Correct option: A

\omega ,{\omega ^2}

Option A (

\omega, \omega^2

) is correct because these are the common roots of both equations, where

\omega

is a complex cube root of unity.

9 Hard

9 of 30

Application +4 | -1

The solution set of the in equation

\left| {2x - 3} \right| < \left| {x + 2} \right|,

View answer and explanation

Correct option: B

\left( {1/3,\;5} \right)

Option B (

\left( {1/3,\;5} \right)

) is correct because it represents the solution set for the given inequality after considering different cases based on the absolute value expressions.

10 Hard

10 of 30

Application +4 | -1

The maximum distance from the origin of coordinates to the point

z

satisfying the equation

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

View answer and explanation

Correct option: C

\frac{1}{2}\left( {\sqrt {{a^2} + 4} + a} \right)

Option C

\frac{{1}}{{2}}\left( {\sqrt {{{{a^2} + 4}}} + a} \right)

is correct because it represents the maximum distance derived by maximizing the modulus of the complex number and solving the resulting quadratic equation.

11 Hard

11 of 30

Analyzing +4 | -1

\left| z \right| = \max \left\{ {\left| {z - 2} \right|,\;\left| {z + 2} \right|} \right\},\;

then

View answer and explanation

Correct option: C

\left| {z + \bar z} \right| = 2

Option C (

\left| {z + \bar z} \right| = 2

) is correct because the given condition implies that the real part of z is either 1 or -1, hence the magnitude of

z + \bar z

is 2.

12 Hard

12 of 30

Analyzing +4 | -1

Suppose

f(x)

is twice differentiable for all real Further suppose

|f(x)| \leq 1

and

\left|f^{\prime \prime}(x)\right| \leq 1

for all rea

x

. Then

\left|f^{\prime}(x)\right|

is
where

f(x+1) =f(x)+\frac{f^{\prime}(x)}{\lfloor}+\frac{f^{\prime}(\xi)}{\lfloor 2}

\text { and } \quad f(x-1) =f(x)-\frac{f^{\prime}(x)}{\lfloor}+\frac{f^{\prime \prime}(\eta)}{\lfloor 2}

for some

\xi

and

\eta

View answer and explanation

Correct option: C

less than or equal to

3 / 2

The correct answer is less than or equal to

3/2

because by using Taylor's expansion and the given conditions, we can bound the value of the first derivative.

13 Hard

13 of 30

Application +4 | -1

a=\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}

, then the value of

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

View answer and explanation

Correct option: B

\frac{(-1)^n}{2^{3 n}}

Option B (

\frac{{(-1)^n}}{{2^{3n}}}

) is correct because after simplification using trigonometric identities and De Moivre's theorem, the expression equals

\frac{{(-1)^n}}{{2^{3n}}}

14 Hard

14 of 30

Application +4 | -1

z

is a complex number satisfying

z + {z^{ - 1}} = 1,

then

{z^n} + {z^{ - n}},n \in N

has the value

View answer and explanation

Correct option: A

2{\left( { - 1} \right)^n},

when

n

is a multiple of 3

Option A (

2{\left( { - 1} \right)^n},

when

n

is a multiple of 3) is correct because solving the given equation yields

z = -\omega

z = -\omega^2

, and substituting these values into

z^n + z^{-n}

gives the stated result when n is a multiple of 3.

15 Hard

15 of 30

Analyzing +4 | -1

{z_1},\;{z_2}

are two complex numbers satisfying

\left| {\frac{{{z_1} + 3{z_2}}}{{3 - {z_1}{{\bar z}_2}}}} \right| = 1,\;\left| {{z_1}} \right| \ne 3

, then

\left| {{z_2}} \right|

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because by manipulating the given equation and using properties of complex numbers, we find that

\left| {{z_2}} \right|

must be equal to 1.

Explore Other Chapters in MATHEMATICS

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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

16 Hard

16 of 30

Analyzing +4 | -1

The joint of

{z_1} = a + i\;b

and

{z_2} = \frac{1}{{ - a + i\;b}}

passes through

View answer and explanation

Correct option: A

Origin

The correct answer is Origin because the equation of the line passing through the given points simplifies to an equation that always passes through the origin.

17 Hard

17 of 30

Analyzing +4 | -1

Let

z

be those complex numbers which satisfy

|z+5| \leq 4

and

z(1+i)+\bar{z}(1-i) \geq-10, i=\sqrt{-1}.

If the maximum value of

|z+1|^2

\alpha+\beta \sqrt{2}

, then the value of

(\alpha+\beta)

is ____. Where

\alpha \& \beta

are natural numbers.

View answer and explanation

Correct option: B

Option B (48) is correct because by substituting

z = x + iy

into the given conditions, we find the region in the complex plane and maximize

|z+1|^2

18 Hard

18 of 30

Analyzing +4 | -1

\sqrt {12 - \sqrt {68 + 48\sqrt 2 } }

is equal to

View answer and explanation

Correct option: C

2 - \sqrt 2

Option C (

2 - \sqrt 2

) is correct because it is the simplified form of the given expression after rationalizing the nested square roots.

19 Hard

19 of 30

Analyzing +4 | -1

a = \cos \left( {\frac{{2\pi }}{7}} \right) + i\sin \left( {\frac{{2\pi }}{7}} \right),\;

then the quadratic equation whose roots are

\mathrm{\alpha } = a + {a^2} + {a^4}\;\mathrm{and}\;\mathrm{\beta } = {a^3} + {a^5} + {a^6}

View answer and explanation

Correct option: D

{x^2} + x + 2 = 0

Option D (

{x^2} + x + 2 = 0

) is correct because calculating the sum and product of the roots α and β leads to this quadratic equation.

20 Hard

20 of 30

Application +4 | -1

When

f(z)

is divided by

z-i

, the remainder is

i

and if

f(z)

is divided by

z+i

, the remainder is

1+i

. If

f(z)

is divided by

z^2+1

, then the remainder is

View answer and explanation

Correct option: B

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

Option B,

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

, is correct because it represents the remainder when f(z) is divided by

z^2+1

, obtained using the remainder theorem.

21 Hard

21 of 30

Application +4 | -1

A\left( {{z_1}} \right),\;B\left( {{z_2}} \right)

and

C\left( {{z_3}} \right)

be the vertices of a triangle

ABC

in which

\angle ABC = \frac{\pi }{4}

and

\frac{{AB}}{{BC}} = \;\sqrt 2

, then the value of

{z_2}

is equal to

View answer and explanation

Correct option: C

{z_3} + i\left( {{z_1} - {z_3}} \right)

Option C (

{z_3} + i\left( {{z_1} - {z_3}} \right)

) is correct because considering the rotation about 'B' and using the given conditions leads to this result.

22 Hard

22 of 30

Analyzing +4 | -1

z \neq 1, \frac{z^2}{z-1}

is real, then point represented by the complex number

z

lies

View answer and explanation

Correct option: B

either on the real axis or on a circle not passing through the origin.

Option B (either on the real axis or on a circle not passing through the origin) is correct because the given condition leads to two possibilities: z lies on the real axis or on a circle centered at (1, 0) with radius 1, which does not pass through the origin.

23 Hard

23 of 30

Application +4 | -1

a=\cos 2 \alpha+i \sin 2 \alpha, \quad b=\cos 2 \beta+i \sin 2 \beta

c=\cos 2 \gamma+i \sin 2 \gamma

and

d=\cos 2\delta+i \sin 2\delta

, then

\sqrt{a b c d}+\frac{1}{\sqrt{a b c d}}

is equal to

View answer and explanation

Correct option: B

2 \cos (\alpha+\beta+\gamma+\delta)

Option B (

2 \cos (\alpha+\beta+\gamma+\delta)

) is correct because by applying De Moivre's theorem and simplifying, the expression equals

2 \cos (\alpha+\beta+\gamma+\delta)

24 Hard

24 of 30

Understanding +4 | -1

f

is as in Question 4, then for

a \in \mathbf{C}

,where

a_1, a_2, a_3

is a permutation of

z_1, z_2, z_3

View answer and explanation

Correct option: A

f\left(z_1, z_2, z_3\right)=f\left(z_1+a, z_2+a, z_3+a\right)

Option A (

f(z_1, z_2, z_3)=f(z_1+a, z_2+a, z_3+a)

) is correct because adding a constant to each variable in the determinant does not change its value due to properties of determinants.

25 Hard

25 of 30

Analyzing +4 | -1

The number of solutions for the equations

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

View answer and explanation

Correct option: A

One solution

Option A (

\frac{1}{2}

) is correct because when the degree of the numerator and denominator are the same, we perform polynomial long division to find the constant term

A

26 Hard

26 of 30

Application +4 | -1

\left| {{a_i}} \right| < 1,{\lambda _i} \ge 0

for

i = 1,2, \ldots .,n

and

{\lambda _1} + {\lambda _2} + \ldots + {\lambda _n} = 1,

then the value of

\left| {{\lambda _1}{a_1} + {\lambda _2}{a_2} + \ldots + {\lambda _n}{a_n}} \right|

View answer and explanation

Correct option: B

Less than 1

Option B (Less than 1) is correct because the magnitude of the weighted sum is less than 1, given the constraints on the magnitudes of a_i and the non-negative weights lambda_i.

27 Hard

27 of 30

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: C

Option C (1) is correct because after simplification using trigonometric identities and Euler's formula, the expression equals 1.

28 Hard

28 of 30

Application +4 | -1

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

are the

n, n

th roots of unity and

z_1, z_2

are any two complex numbers, then

\sum_{k=0}^{n-1}\left|z_1+\omega^k z_2\right|^2

is equal to

View answer and explanation

Correct option: A

n\left[\left|z_1\right|^2+\left|z_2\right|^2\right]

Option A (

n[|z_1|^2+|z_2|^2]

) is correct because it is derived from expanding the sum and using the properties of nth roots of unity.

29 Hard

29 of 30

Analyzing +4 | -1

Let

a, b

be two real numbers such that

a b < 0

. If the complex number

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

is of unit modulus and

a+

i b

lies on the circle

|z-1|=|2 z|

, then a possible value of

\frac{1+[a]}{4 b}

, where

[t]

is greatest integer function is:

View answer and explanation

Correct option: C

-\frac{1}{2}

The correct answer is C (

-\frac{1}{2}

) because it is obtained by using the given conditions to find the values of a and b, and then evaluating the expression.

30 Hard

30 of 30

Application +4 | -1

2 \cos \theta=x+\frac{1}{x}

and

2 \cos \phi=y+\frac{1}{y}

, then

View answer and explanation

Correct option: B

x^m y^n+\frac{1}{x^m y^n}=2 \cos (m \theta+n \phi)

Option B (

x^m y^n+\frac{{1}}{{x^m y^n}}=2 \cos (m \theta+n \phi)

) is correct because it follows from De Moivre's theorem.

Revision notes and key takeaways

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

Key Takeaways

✓ This set contains 30 single-correct-answer MCQs on COMPLEX NUMBERS.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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