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Home/JEE MAINS/mathematics/complex numbers
Curated PYQ Collection

Top 50 Most Repeated COMPLEX NUMBERS PYQs | JEE MAINS

A curated collection of the most important questions from COMPLEX NUMBERS, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

A.-1+2i
B.1-2i
C.-2- i
D.2+ i

Concept Applied

Multiply numerator and denominator by the conjugate of the denominator: \(\frac{3+4i}{1-2i}\times\frac{1+2i}{1+2i}=\frac{(3+4i)(1+2i)}{1^2+2^2}=\frac{...

Read Full Step-by-Step Solution β†’

Question #2

Practice Question

A.$\frac{3}{25} - \frac{4}{25}i$
B.$\frac{3}{7} - \frac{4}{7}i$
C.$\frac{1}{3} + \frac{1}{4}i$
D.$\frac{3}{25} + \frac{4}{25}i$

Concept Applied

Multiply numerator and denominator by conjugate: $\frac{1}{3+4i} \cdot \frac{3-4i}{3-4i} = \frac{3-4i}{25}$...

Read Full Step-by-Step Solution β†’

Question #3

Practice Question

A.$-8$
B.$0$
C.$8$
D.$-1$

Concept Applied

Using

+ \omega + \omega^2 = 0$, we get
+ \omega = -\omega^2$, so
+ \omega - \omega^2 = -2\omega^2$. Then $(-2\omega^2)^3 = -8\omega^6 = -8(1)...

Read Full Step-by-Step Solution β†’

Question #4

Practice Question

Concept Applied

Points lie on unit circle and centroid at origin β†’ circumcenter is 0....

Read Full Step-by-Step Solution β†’

Question #5

Practice Question

A.A circle with center at $(0, 2)$ and radius 2
B.A circle with center at $(2, 0)$ and radius 2
C.A straight line parallel to the real axis
D.A circle with center at $(0, -2)$ and radius 2

Concept Applied

The equation $|z - 2i| = 2$ represents all points $z$ whose distance from

...

Read Full Step-by-Step Solution β†’

Question #6

Practice Question

Concept Applied

Apply De Moivre's theorem using polar form....

Read Full Step-by-Step Solution β†’

Question #7

Practice Question

A.$\frac{7 + 24i}{25}$
B.$\frac{24 + 7i}{25}$
C.$\frac{-7 + 24i}{25}$
D.$\frac{7 - 24i}{25}$

Concept Applied

$\frac{z}{\bar{z}} = \frac{(3+4i)^2}{(3-4i)(3+4i)} = \frac{-7+24i}{25}$ using conjugate division....

Read Full Step-by-Step Solution β†’

Question #8

Practice Question

A.2
B.\sqrt{2}
C.\sqrt{3}
D.4

Concept Applied

For a complex number $z = a + ib$, the modulus is $|z| = \sqrt{a^2 + b^2}$. Here $a = 1$ and $b = \sqrt{3}$, so $|z| = \sqrt{1^2 + (\sqrt{3})^2} = \sq...

Read Full Step-by-Step Solution β†’

Question #9

Practice Question

A.0
B.1
C.-1
D.$\omega$

Concept Applied

Using

+\omega+\omega^2=0$ and $\omega^3=1$, the product simplifies to 1....

Read Full Step-by-Step Solution β†’

Question #10

Practice Question

A.11 - 2i
B.5 + 2i
C.11 + 2i
D.-5 + 10i

Concept Applied

$(3+4i)(1-2i) = 3(1) + 3(-2i) + 4i(1) + 4i(-2i) = 3 -6i +4i +8 = 11 -2i$...

Read Full Step-by-Step Solution β†’

Question #11

Practice Question

Concept Applied

Step 1: We need to find the value of $z^4$. Step 2: We can use the equation of $z$ to find the value of $z^4$. Step 3: The equation of $z$ is given by...

Read Full Step-by-Step Solution β†’

Question #12

Practice Question

A.Theoretical foundations
B.Practical applications
C.Experimental data
D.Historical context

Concept Applied

Foundational check for Complex Numbers in Class 11. Study the core principles carefully for competitive exams....

Read Full Step-by-Step Solution β†’

Question #13

Practice Question

A.-1+2i
B.1-2i
C.-2- i
D.2+ i

Concept Applied

Multiply numerator and denominator by the conjugate of the denominator: \(\frac{3+4i}{1-2i}\times\frac{1+2i}{1+2i}=\frac{(3+4i)(1+2i)}{1^2+2^2}=\frac{...

Read Full Step-by-Step Solution β†’

Question #14

Practice Question

A.$\frac{3}{25} - \frac{4}{25}i$
B.$\frac{3}{7} - \frac{4}{7}i$
C.$\frac{1}{3} + \frac{1}{4}i$
D.$\frac{3}{25} + \frac{4}{25}i$

Concept Applied

Multiply numerator and denominator by conjugate: $\frac{1}{3+4i} \cdot \frac{3-4i}{3-4i} = \frac{3-4i}{25}$...

Read Full Step-by-Step Solution β†’

Question #15

Practice Question

A.$-8$
B.$0$
C.$8$
D.$-1$

Concept Applied

Using

+ \omega + \omega^2 = 0$, we get
+ \omega = -\omega^2$, so
+ \omega - \omega^2 = -2\omega^2$. Then $(-2\omega^2)^3 = -8\omega^6 = -8(1)...

Read Full Step-by-Step Solution β†’

Question #16

Practice Question

Concept Applied

Points lie on unit circle and centroid at origin β†’ circumcenter is 0....

Read Full Step-by-Step Solution β†’

Question #17

Practice Question

A.A circle with center at $(0, 2)$ and radius 2
B.A circle with center at $(2, 0)$ and radius 2
C.A straight line parallel to the real axis
D.A circle with center at $(0, -2)$ and radius 2

Concept Applied

The equation $|z - 2i| = 2$ represents all points $z$ whose distance from

...

Read Full Step-by-Step Solution β†’

Question #18

Practice Question

Concept Applied

Apply De Moivre's theorem using polar form....

Read Full Step-by-Step Solution β†’

Question #19

Practice Question

A.$\frac{7 + 24i}{25}$
B.$\frac{24 + 7i}{25}$
C.$\frac{-7 + 24i}{25}$
D.$\frac{7 - 24i}{25}$

Concept Applied

$\frac{z}{\bar{z}} = \frac{(3+4i)^2}{(3-4i)(3+4i)} = \frac{-7+24i}{25}$ using conjugate division....

Read Full Step-by-Step Solution β†’

Question #20

Practice Question

A.2
B.\sqrt{2}
C.\sqrt{3}
D.4

Concept Applied

For a complex number $z = a + ib$, the modulus is $|z| = \sqrt{a^2 + b^2}$. Here $a = 1$ and $b = \sqrt{3}$, so $|z| = \sqrt{1^2 + (\sqrt{3})^2} = \sq...

Read Full Step-by-Step Solution β†’

Question #21

Practice Question

A.0
B.1
C.-1
D.$\omega$

Concept Applied

Using

+\omega+\omega^2=0$ and $\omega^3=1$, the product simplifies to 1....

Read Full Step-by-Step Solution β†’

Question #22

Practice Question

A.11 - 2i
B.5 + 2i
C.11 + 2i
D.-5 + 10i

Concept Applied

$(3+4i)(1-2i) = 3(1) + 3(-2i) + 4i(1) + 4i(-2i) = 3 -6i +4i +8 = 11 -2i$...

Read Full Step-by-Step Solution β†’

Question #23

Practice Question

Concept Applied

Step 1: We need to find the value of $z^4$. Step 2: We can use the equation of $z$ to find the value of $z^4$. Step 3: The equation of $z$ is given by...

Read Full Step-by-Step Solution β†’

Question #24

Practice Question

A.Theoretical foundations
B.Practical applications
C.Experimental data
D.Historical context

Concept Applied

Foundational check for Complex Numbers in Class 11. Study the core principles carefully for competitive exams....

Read Full Step-by-Step Solution β†’

Question #25

Practice Question

A.-1+2i
B.1-2i
C.-2- i
D.2+ i

Concept Applied

Multiply numerator and denominator by the conjugate of the denominator: \(\frac{3+4i}{1-2i}\times\frac{1+2i}{1+2i}=\frac{(3+4i)(1+2i)}{1^2+2^2}=\frac{...

Read Full Step-by-Step Solution β†’

Question #26

Practice Question

A.$\frac{3}{25} - \frac{4}{25}i$
B.$\frac{3}{7} - \frac{4}{7}i$
C.$\frac{1}{3} + \frac{1}{4}i$
D.$\frac{3}{25} + \frac{4}{25}i$

Concept Applied

Multiply numerator and denominator by conjugate: $\frac{1}{3+4i} \cdot \frac{3-4i}{3-4i} = \frac{3-4i}{25}$...

Read Full Step-by-Step Solution β†’

Question #27

Practice Question

A.$-8$
B.$0$
C.$8$
D.$-1$

Concept Applied

Using

+ \omega + \omega^2 = 0$, we get
+ \omega = -\omega^2$, so
+ \omega - \omega^2 = -2\omega^2$. Then $(-2\omega^2)^3 = -8\omega^6 = -8(1)...

Read Full Step-by-Step Solution β†’

Question #28

Practice Question

Concept Applied

Points lie on unit circle and centroid at origin β†’ circumcenter is 0....

Read Full Step-by-Step Solution β†’

Question #29

Practice Question

A.A circle with center at $(0, 2)$ and radius 2
B.A circle with center at $(2, 0)$ and radius 2
C.A straight line parallel to the real axis
D.A circle with center at $(0, -2)$ and radius 2

Concept Applied

The equation $|z - 2i| = 2$ represents all points $z$ whose distance from

...

Read Full Step-by-Step Solution β†’

Question #30

Practice Question

Concept Applied

Apply De Moivre's theorem using polar form....

Read Full Step-by-Step Solution β†’

Question #31

Practice Question

A.$\frac{7 + 24i}{25}$
B.$\frac{24 + 7i}{25}$
C.$\frac{-7 + 24i}{25}$
D.$\frac{7 - 24i}{25}$

Concept Applied

$\frac{z}{\bar{z}} = \frac{(3+4i)^2}{(3-4i)(3+4i)} = \frac{-7+24i}{25}$ using conjugate division....

Read Full Step-by-Step Solution β†’

Question #32

Practice Question

A.2
B.\sqrt{2}
C.\sqrt{3}
D.4

Concept Applied

For a complex number $z = a + ib$, the modulus is $|z| = \sqrt{a^2 + b^2}$. Here $a = 1$ and $b = \sqrt{3}$, so $|z| = \sqrt{1^2 + (\sqrt{3})^2} = \sq...

Read Full Step-by-Step Solution β†’

Question #33

Practice Question

A.0
B.1
C.-1
D.$\omega$

Concept Applied

Using

+\omega+\omega^2=0$ and $\omega^3=1$, the product simplifies to 1....

Read Full Step-by-Step Solution β†’

Question #34

Practice Question

A.11 - 2i
B.5 + 2i
C.11 + 2i
D.-5 + 10i

Concept Applied

$(3+4i)(1-2i) = 3(1) + 3(-2i) + 4i(1) + 4i(-2i) = 3 -6i +4i +8 = 11 -2i$...

Read Full Step-by-Step Solution β†’

Question #35

Practice Question

Concept Applied

Step 1: We need to find the value of $z^4$. Step 2: We can use the equation of $z$ to find the value of $z^4$. Step 3: The equation of $z$ is given by...

Read Full Step-by-Step Solution β†’

Question #36

Practice Question

A.Theoretical foundations
B.Practical applications
C.Experimental data
D.Historical context

Concept Applied

Foundational check for Complex Numbers in Class 11. Study the core principles carefully for competitive exams....

Read Full Step-by-Step Solution β†’

Question #37

Practice Question

A.-1+2i
B.1-2i
C.-2- i
D.2+ i

Concept Applied

Multiply numerator and denominator by the conjugate of the denominator: \(\frac{3+4i}{1-2i}\times\frac{1+2i}{1+2i}=\frac{(3+4i)(1+2i)}{1^2+2^2}=\frac{...

Read Full Step-by-Step Solution β†’

Question #38

Practice Question

A.$\frac{3}{25} - \frac{4}{25}i$
B.$\frac{3}{7} - \frac{4}{7}i$
C.$\frac{1}{3} + \frac{1}{4}i$
D.$\frac{3}{25} + \frac{4}{25}i$

Concept Applied

Multiply numerator and denominator by conjugate: $\frac{1}{3+4i} \cdot \frac{3-4i}{3-4i} = \frac{3-4i}{25}$...

Read Full Step-by-Step Solution β†’

Question #39

Practice Question

A.$-8$
B.$0$
C.$8$
D.$-1$

Concept Applied

Using

+ \omega + \omega^2 = 0$, we get
+ \omega = -\omega^2$, so
+ \omega - \omega^2 = -2\omega^2$. Then $(-2\omega^2)^3 = -8\omega^6 = -8(1)...

Read Full Step-by-Step Solution β†’

Question #40

Practice Question

Concept Applied

Points lie on unit circle and centroid at origin β†’ circumcenter is 0....

Read Full Step-by-Step Solution β†’

Question #41

Practice Question

A.A circle with center at $(0, 2)$ and radius 2
B.A circle with center at $(2, 0)$ and radius 2
C.A straight line parallel to the real axis
D.A circle with center at $(0, -2)$ and radius 2

Concept Applied

The equation $|z - 2i| = 2$ represents all points $z$ whose distance from

...

Read Full Step-by-Step Solution β†’

Question #42

Practice Question

Concept Applied

Apply De Moivre's theorem using polar form....

Read Full Step-by-Step Solution β†’

Question #43

Practice Question

A.$\frac{7 + 24i}{25}$
B.$\frac{24 + 7i}{25}$
C.$\frac{-7 + 24i}{25}$
D.$\frac{7 - 24i}{25}$

Concept Applied

$\frac{z}{\bar{z}} = \frac{(3+4i)^2}{(3-4i)(3+4i)} = \frac{-7+24i}{25}$ using conjugate division....

Read Full Step-by-Step Solution β†’

Question #44

Practice Question

A.2
B.\sqrt{2}
C.\sqrt{3}
D.4

Concept Applied

For a complex number $z = a + ib$, the modulus is $|z| = \sqrt{a^2 + b^2}$. Here $a = 1$ and $b = \sqrt{3}$, so $|z| = \sqrt{1^2 + (\sqrt{3})^2} = \sq...

Read Full Step-by-Step Solution β†’

Question #45

Practice Question

A.0
B.1
C.-1
D.$\omega$

Concept Applied

Using

+\omega+\omega^2=0$ and $\omega^3=1$, the product simplifies to 1....

Read Full Step-by-Step Solution β†’

Question #46

Practice Question

A.11 - 2i
B.5 + 2i
C.11 + 2i
D.-5 + 10i

Concept Applied

$(3+4i)(1-2i) = 3(1) + 3(-2i) + 4i(1) + 4i(-2i) = 3 -6i +4i +8 = 11 -2i$...

Read Full Step-by-Step Solution β†’

Question #47

Practice Question

Concept Applied

Step 1: We need to find the value of $z^4$. Step 2: We can use the equation of $z$ to find the value of $z^4$. Step 3: The equation of $z$ is given by...

Read Full Step-by-Step Solution β†’

Question #48

Practice Question

A.Theoretical foundations
B.Practical applications
C.Experimental data
D.Historical context

Concept Applied

Foundational check for Complex Numbers in Class 11. Study the core principles carefully for competitive exams....

Read Full Step-by-Step Solution β†’

Question #49

Practice Question

A.-1+2i
B.1-2i
C.-2- i
D.2+ i

Concept Applied

Multiply numerator and denominator by the conjugate of the denominator: \(\frac{3+4i}{1-2i}\times\frac{1+2i}{1+2i}=\frac{(3+4i)(1+2i)}{1^2+2^2}=\frac{...

Read Full Step-by-Step Solution β†’

Question #50

Practice Question

A.$\frac{3}{25} - \frac{4}{25}i$
B.$\frac{3}{7} - \frac{4}{7}i$
C.$\frac{1}{3} + \frac{1}{4}i$
D.$\frac{3}{25} + \frac{4}{25}i$

Concept Applied

Multiply numerator and denominator by conjugate: $\frac{1}{3+4i} \cdot \frac{3-4i}{3-4i} = \frac{3-4i}{25}$...

Read Full Step-by-Step Solution β†’
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