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Class 11 RD Sharma Solutions - Chapter 13 Complex Numbers - Exercise 13.2 | Set 2

Last Updated : 23 Jul, 2025

In this article, we will be going to solve the entire exercise 13.2 of our NCERT textbook. Complex numbers are an extension of real numbers and are fundamental in various fields of mathematics, engineering, and physics. They provide a way to solve equations that have no real solutions, such as x2+1=0. Complex numbers also allow for a more comprehensive understanding of polynomial equations and play a crucial role in advanced mathematical theories and applications.

What are Complex Numbers?

A complex number is a number of the form z = a+bi, where:

  • a and b are real numbers.
  • i is the imaginary unit, defined by the property i2 = -1.

In this form, 'a' is called the real part of the complex number z, and 'b' is called the imaginary part.

Chapter 13 Complex Numbers - Exercise 13.2 | Set 2 Solutions

Let's learn in detail about, solutions to NCERT Exercise - 13.2| Set 2 chapter 13 Complex Number in the article below:

Question 14. If, find (a, b).

Solution:

We have,

ā‡’

ā‡’

ā‡’

ā‡’

ā‡’ (āˆ’i)100 = a + ib

ā‡’ a + ib = 1

On comparing real and imaginary parts on both sides, we get,

ā‡’ (a, b) = (1, 0)

Question 15. If a = cos Īø + i sin Īø, find the value of.

Solution:

Given a = cos Īø + i sin Īø, we get,

=

=

=

=

=

=

=

=

=

=

Therefore, the value ofis.

Question 16. Evaluate the following :

(i) 2x3 + 2x2 āˆ’ 7x + 72, when x = (3āˆ’5i)/2

Solution:

We have, x = (3āˆ’5i)/2

ā‡’ 2x = 3 āˆ’ 5i

ā‡’ 2x āˆ’ 3 = āˆ’5i

ā‡’ (2x āˆ’ 3)2 = 25i2

ā‡’ 4x2 + 9 āˆ’ 12x = āˆ’25

ā‡’ 4x2 āˆ’ 12x + 34 = 0

ā‡’ 2x2 āˆ’ 6x + 17 = 0

Now, 2x3 + 2x2 āˆ’ 7x + 72 = x (2x2 āˆ’ 6x + 17) + 6x2 āˆ’ 17x + 2x2 āˆ’ 7x + 72

= x (0) + 8x2 āˆ’ 24x + 72

= 4 (2x2 āˆ’ 6x + 17) + 4

= 4 (0) + 4

= 4

Therefore, the value of 2x3 + 2x2 āˆ’ 7x + 72 is 4.

(ii) x4 āˆ’ 4x3 + 4x2 +8x +44, when x = 3 + 2i

Solution:

We have, x = 3 + 2i

ā‡’ x āˆ’ 3 = 2i

ā‡’ (x āˆ’ 3)2 = (2i)2

ā‡’ x2 + 9 āˆ’ 6x = 4i2

ā‡’ x2 āˆ’ 6x + 9 + 4 = 0

ā‡’ x2 āˆ’ 6x + 13 = 0

Now, x4 āˆ’ 4x3 + 4x2 + 8x + 44 = x2 (x2 āˆ’ 6x + 13) + 6x3 āˆ’ 13x2 āˆ’ 4x3 + 4x2 + 8x + 44

= 2x3 āˆ’ 9x2 + 8x + 44

= 2x (x2 āˆ’ 6x + 13) + 12x2 āˆ’ 26x āˆ’ 9x2 + 8x + 44

= 3x2 āˆ’ 18x + 44

= 3 (x2 āˆ’ 6x + 13) + 5

= 5

Therefore, the value of x4 āˆ’ 4x3 + 4x2 + 8x + 44 is 5.

(iii) x4 + 4x3 + 6x2 + 4x + 9, when x = āˆ’1 + iāˆš2

Solution:

We have, x = āˆ’1 + iāˆš2

ā‡’ x + 1 = iāˆš2

ā‡’ (x + 1)2 = 2i2

ā‡’ x2 + 1 + 2x = āˆ’2

ā‡’ x2 + 2x + 3 = 0

Now, x4 + 4x3 + 6x2 + 4x + 9 = x2 (x2 + 2x + 3) āˆ’ 2x3 āˆ’ 3x2 + 4x3 + 6x2 + 4x + 9

= 2x3 + 3x2 + 4x + 9

= 2x (x2 + 2x + 3) āˆ’ 4x2 āˆ’ 6x + 3x2 + 4x + 9

= āˆ’ x2 āˆ’ 2x + 9

= āˆ’ (x2 + 2x + 3) + 3 + 9

= 3 + 9

= 12

Therefore, the value of x4 + 4x3 + 6x2 + 4x + 9 is 12.

(iv) x6 + x4 + x2 + 1, when x = (1+i)/āˆš2

Solution:

We have, x = (1+i)/āˆš2

ā‡’ āˆš2x = 1 + i

ā‡’ 2x2 = 1 + i2 + 2i

ā‡’ 2x2 = 2i

ā‡’ 4x4 = 4i2

ā‡’ x4 = āˆ’1

ā‡’ x4 + 1 = 0

Now, x6 + x4 + x2 + 1 = (x6 + x2) + (x4 +1)

= x6 + x2

= x2 (x4 + 1)

= 0

Therefore, the value of x6 + x4 + x2 + 1 is 0.

(v) 2x4 + 5x3 + 7x2 āˆ’ x + 41, when x = āˆ’2 āˆ’ āˆš3i

Solution:

We have, x = āˆ’2 āˆ’ āˆš3i

x2 = (āˆ’2 āˆ’ āˆš3i)2 = 4 + 4āˆš3i + 3i2 = 1 + 4āˆš3i

x3 = (1 + 4āˆš3i) (āˆ’2 āˆ’ āˆš3i) = āˆ’2 āˆ’ āˆš3i āˆ’ 8āˆš3i āˆ’12i2 = 10 āˆ’ 9āˆš3i

x4 = (1 + 4āˆš3i)2 = 1 + 8āˆš3i + 48i2 = āˆ’47 + 8āˆš3i

Now, 2x4 + 5x3 + 7x2 āˆ’ x + 41 becomes,

= 2(āˆ’47 + 8āˆš3i) + 5(10 āˆ’ 9āˆš3i) + 7(1 + 4āˆš3i) āˆ’ (āˆ’2 āˆ’ āˆš3i) + 41

= āˆ’94 + 16āˆš3i + 50 āˆ’ 45āˆš3i + 7 + 28āˆš3i + 2 + āˆš3i + 41

= 6

Therefore, the value of 2x4 + 5x3 + 7x2 āˆ’ x + 41 is 6.

Question 17. For a positive integer n, find the value of (1āˆ’i)n (1āˆ’1/i)n.

Solution:

We have,

(1āˆ’i)n (1āˆ’1/i)n = (1āˆ’i)n

=

=

=

= 2n

Therefore, the value of (1āˆ’i)n (1āˆ’1/i)n is 2n.

Question 18. If (1+i)z = (1āˆ’i), then show that z = āˆ’i.

Solution:

We have,

ā‡’ (1+i)z = (1āˆ’i)

ā‡’ z =

ā‡’ z =

ā‡’ z =

ā‡’ z =

ā‡’ z = āˆ’i

Hence proved.

Question 19. Solve the system of equations: Re(z2) = 0, |z| = 2.

Solution:

Let z = x + iy.

Now z2 = (x + iy)2

= x2 + i2y2 + 2xyi

= x2 āˆ’ y2 + 2xyi

We have, Re(z2) = 0

ā‡’ x2 āˆ’ y2 = 0 . . . . (1)

Also, it is given, |z| = 2.

ā‡’= 2

ā‡’ x2 + y2 = 4 . . . . (2)

Solving (1) and (2), we get, x = Ā±āˆš2 and y = Ā±āˆš2.

Therefore, x + iy = Ā±āˆš2 Ā± āˆš2i .

Question 20. Ifis purely imaginary number (zā‰ āˆ’1), find the value of |z|.

Solution:

Let z = x + iy

We have,

=

=

=

=

=

As the complex number is purely imaginary, therefore,

ā‡’ Re(z) = 0

ā‡’= 0

ā‡’ x2 + y2 = 1

ā‡’= 1

ā‡’ |z| = 1

Therefore, the value of |z| is 1.

Question 21. If z1 is a complex number other than āˆ’1 such that |z1| = 1 and z2 =,then show that the real parts of z2 is zero.

Solution:

Given |z| = 1

ā‡’ |z|2 = 1

ā‡’ x2 + y2 = 1 . . . . (1)

Let z1 = x + iy and z2 = a + ib.

According to the question, we have,

ā‡’ z2 =

ā‡’ a + ib =

ā‡’ a + ib =

ā‡’ a + ib =

ā‡’ a + ib =

Using (1) we get,

ā‡’ a + ib =

ā‡’ a + ib =

On comparing the real and imaginary parts on both sides, we get a = 0.

Therefore, the real parts of z2 is 0. Hence proved.

Question 22. If |z+1| = z + 2(1+i), find z.

Solution:

Let z = x + iy. According to the question, we have,

ā‡’ |x + iy + 1| = x + iy + 2(1 + i)

ā‡’= (x + 2) + i(y + 2)

On comparing the real and imaginary parts, we get

ā‡’ y + 2 = 0

ā‡’ y = āˆ’2

And also,

ā‡’ x + 2 =

ā‡’ (x + 2)2 = (x+1)2 + y2

ā‡’ x2 + 4 + 4x = x2 + 2x + 1+ y2

ā‡’ 2x = y2 āˆ’ 3

ā‡’ 2x = 4 āˆ’ 3

ā‡’ 2x = 1

ā‡’ x = 1/2

Therefore, z = x + iy = 1/2 āˆ’2i.

Question 23. Solve the equation: |z| = z + 1 + 2i.

Solution:

Let z = x + iy. According to the question, we have,

ā‡’ |z| = z + 1 + 2i

ā‡’ |x + iy| = x + iy + 1 + 2i

ā‡’= (x + 1) + (y + 2)i

ā‡’ x2 + y2 = (x+1)2 + (y+2)2i2 + 2 (x+1) (y+2)i

ā‡’ x2 + y2 = x2+1 + 2x āˆ’ y2 āˆ’ 1 + 2y + 2 (x+1) (y+2)i

ā‡’ 2y2 āˆ’ 2x + 4y + 4 = 2i (x+1) (y+2)

ā‡’ y2 āˆ’ x + 2y + 2 = i (x+1) (y+2)

On comparing both sides, we get,

ā‡’ (x+1) (y+2) = 0

ā‡’ x = āˆ’1 and y = āˆ’2

Also, y2 āˆ’ x + 2y + 2 = 0

Taking x = āˆ’1, we get y2 āˆ’ (āˆ’1) + 2y + 2 = 0

ā‡’ y2 + 2y + 3 = 0, which doesn't have a solution as the roots are imaginary.

Taking y = āˆ’2, (4 āˆ’ x āˆ’4 + 2) = 0

ā‡’ x = 2

Therefore, z = x + iy = 2 āˆ’ 2i.

Question 24. What is the smallest positive integer n for which (1+i)2n = (1āˆ’i)2n?

Solution:

We are given,

ā‡’ (1+i)2n = (1āˆ’i)2n

ā‡’= 1

ā‡’= 1

ā‡’= 1

ā‡’= 1

ā‡’ i2n = 1

ā‡’ i2n = i4

ā‡’ 2n = 4

ā‡’ n = 2

Therefore, the smallest positive integer n for which (1+i)2n = (1āˆ’i)2n is 2.

Question 25. If z1, z2, z3 are complex numbers such that |z1| = |z2| = |z3| == 1, then find the value of |z1 + z2 + z3|.

Solution:

We are given,

|z1| = |z2| = |z3| == 1

Now, |z1 + z2 + z3| =

=

=

= 1

Therefore, the value of |z1 + z2 + z3| is 1.

Question 26. Find the number of solutions of z2 + |z|2 = 0.

Solution:

Let z = x + iy. We have,

ā‡’ z2 + |z|2 = 0

ā‡’ (x + iy)2 + |x + iy|2 = 0

ā‡’ x2 + i2y2 + 2xyi + x2 + y2 = 0

ā‡’ x2 āˆ’ y2 + 2xyi + x2 + y2 = 0

ā‡’ 2x2 + 2xyi = 0

On comparing the real and imaginary parts on both sides, we get

ā‡’ 2x2 = 0 and 2xy = 0

ā‡’ x = 0 and y āˆˆ R

Therefore, z = 0 + iy, where y āˆˆ R.

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