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

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\left(\frac{1-i}{1+i}\right)^{100}=a+ib , find (a, b).

Solution:

We have,

⇒\left(\frac{1-i}{1+i}\right)^{100}=a+ib

⇒\left[\frac{(1-i)^2}{(1+i)(1-i)}\right]^{100}=a+ib

⇒\left(\frac{1+i^2-2i}{1-i^2}\right)^{100}=a+ib

⇒\left(\frac{-2i}{2}\right)^{100}=a+ib

⇒ (−i)¹⁰⁰ = a + ib

⇒ a + ib = 1

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

⇒ (a, b) = (1, 0)

Question 15. If a = cos θ + i sin θ, find the value of\frac{1+a}{1-a} .

Solution:

Given a = cos θ + i sin θ, we get,

\frac{1+a}{1-a} =\frac{1+cosθ+isinθ}{1-cosθ-isinθ}

=\frac{(1+cosθ+isinθ)(1-cosθ+isinθ)}{(1-cosθ-isinθ)(1-cosθ+isinθ)}

=\frac{(1+isinθ)^2-cos^2θ}{(1-cosθ)^2-i^2sin^2θ}

=\frac{1+i^2sin^2θ+2isinθ-cos^2θ}{1+cos^2θ-2cosθ-i^2sin^2θ}

=\frac{1+2isinθ-sin^2θ-cos^2θ}{1-2cosθ+cos^2θ+sin^2θ}

=\frac{1+2isinθ-1}{1-2cosθ+1}

=\frac{2isinθ}{2-2cosθ}

=\frac{isinθ}{1-cosθ}

=\frac{2isin\frac{θ}{2}cos\frac{θ}{2}}{2sin^2\frac{θ}{2}}

=icot\frac{θ}{2}

Therefore, the value of\frac{1+a}{1-a} isicot\frac{θ}{2} .

Question 16. Evaluate the following :

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

Solution:

We have, x = (3−5i)/2

⇒ 2x = 3 − 5i

⇒ 2x − 3 = −5i

⇒ (2x − 3)² = 25i²

⇒ 4x² + 9 − 12x = −25

⇒ 4x² − 12x + 34 = 0

⇒ 2x² − 6x + 17 = 0

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

= x (0) + 8x² − 24x + 72

= 4 (2x² − 6x + 17) + 4

= 4 (0) + 4

= 4

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

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

Solution:

We have, x = 3 + 2i

⇒ x − 3 = 2i

⇒ (x − 3)² = (2i)²

⇒ x² + 9 − 6x = 4i²

⇒ x² − 6x + 9 + 4 = 0

⇒ x² − 6x + 13 = 0

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

= 2x³ − 9x² + 8x + 44

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

= 3x² − 18x + 44

= 3 (x² − 6x + 13) + 5

= 5

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

(iii) x⁴ + 4x³ + 6x² + 4x + 9, when x = −1 + i√2

Solution:

We have, x = −1 + i√2

⇒ x + 1 = i√2

⇒ (x + 1)² = 2i²

⇒ x² + 1 + 2x = −2

⇒ x² + 2x + 3 = 0

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

= 2x³ + 3x² + 4x + 9

= 2x (x² + 2x + 3) − 4x² − 6x + 3x² + 4x + 9

= − x² − 2x + 9

= − (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)/√2

Solution:

We have, x = (1+i)/√2

⇒ √2x = 1 + i

⇒ 2x² = 1 + i² + 2i

⇒ 2x² = 2i

⇒ 4x⁴ = 4i²

⇒ x4 = −1

⇒ 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 − x + 41, when x = −2 − √3i

Solution:

We have, x = −2 − √3i

x² = (−2 − √3i)² = 4 + 4√3i + 3i² = 1 + 4√3i

x³ = (1 + 4√3i) (−2 − √3i) = −2 − √3i − 8√3i −12i² = 10 − 9√3i

x⁴ = (1 + 4√3i)² = 1 + 8√3i + 48i² = −47 + 8√3i

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

= 2(−47 + 8√3i) + 5(10 − 9√3i) + 7(1 + 4√3i) − (−2 − √3i) + 41

= −94 + 16√3i + 50 − 45√3i + 7 + 28√3i + 2 + √3i + 41

= 6

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

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

Solution:

We have,

(1−i)ⁿ (1−1/i)ⁿ = (1−i)ⁿ(\frac{i-1}{i})^n

=\left[\frac{(1-i)^2}{-i}\right]^n

=\left(\frac{1+i^2-2i}{-i}\right)^n

=\left(\frac{-2i}{-i}\right)^n

= 2ⁿ

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

Question 18. If (1+i)z = (1−i)\bar{z} , then show that z = −i\bar{z} .

Solution:

We have,

⇒ (1+i)z = (1−i)\bar{z}

⇒ z =(\frac{1-i}{1+i})\bar{z}

⇒ z =\left[\frac{(1-i)^2}{(1+i)(1-i)}\right]\bar{z}

⇒ z =\left[\frac{1+i^2-2i}{1-i^2}\right]\bar{z}

⇒ z =\left[\frac{-2i}{2}\right]\bar{z}

⇒ z = −i\bar{z}

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 − y² + 2xyi

We have, Re(z²) = 0

⇒ x² − y² = 0 . . . . (1)

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

⇒\sqrt{x^2+y^2} = 2

⇒ x² + y² = 4 . . . . (2)

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

Therefore, x + iy = ±√2 ± √2i .

Question 20. If\frac{z-1}{z+1} is purely imaginary number (z≠−1), find the value of |z|.

Solution:

Let z = x + iy

We have,\frac{z-1}{z+1}

=\frac{x-1+iy}{x+1+iy}

=\frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}

=\frac{x^2+x-ixy-x-1+iy+ixy+iy+y^2}{(x+1)^2-(iy)^2}

=\frac{x^2+y^2-1+2iy}{x^2+y^2+1+2xy}

=\frac{x^2+y^2-1}{x^2+y^2+1+2xy}+\frac{2yi}{x^2+y^2+1+2xy}

As the complex number is purely imaginary, therefore,

⇒ Re(z) = 0

⇒\frac{x^2+y^2-1}{x^2+y^2+1+2xy} = 0

⇒ x² + y² = 1

⇒\sqrt{x^2+y^2} = 1

⇒ |z| = 1

Therefore, the value of |z| is 1.

Question 21. If z₁ is a complex number other than −1 such that |z₁| = 1 and z₂ =\frac{z_1-1}{z_1+1} ,then show that the real parts of z₂ is zero.

Solution:

Given |z| = 1

⇒ |z|² = 1

⇒ x² + y² = 1 . . . . (1)

Let z₁ = x + iy and z₂ = a + ib.

According to the question, we have,

⇒ z₂ =\frac{z_1-1}{z_1+1}

⇒ a + ib =\frac{x+iy-1}{x+iy+1}

⇒ a + ib =\frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}

⇒ a + ib =\frac{x^2-1+y^2-iyx+iy+iyx+iy}{(x+1)^2-(iy)^2}

⇒ a + ib =\frac{(x^2+y^2)-1+2iy}{(x+1)^2-(iy)^2}

Using (1) we get,

⇒ a + ib =\frac{1-1+2iy}{(x+1)^2-(iy)^2}

⇒ a + ib =\frac{2iy}{(x+1)^2-(iy)^2}

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,

⇒ |x + iy + 1| = x + iy + 2(1 + i)

⇒\sqrt{(x+1)^2+y^2} = (x + 2) + i(y + 2)

On comparing the real and imaginary parts, we get

⇒ y + 2 = 0

⇒ y = −2

And also,

⇒ x + 2 =\sqrt{(x+1)^2+y^2}

⇒ (x + 2)² = (x+1)² + y²

⇒ x² + 4 + 4x = x² + 2x + 1+ y²

⇒ 2x = y² − 3

⇒ 2x = 4 − 3

⇒ 2x = 1

⇒ x = 1/2

Therefore, z = x + iy = 1/2 −2i.

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

Solution:

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

⇒ |z| = z + 1 + 2i

⇒ |x + iy| = x + iy + 1 + 2i

⇒\sqrt{x^2+y^2} = (x + 1) + (y + 2)i

⇒ x² + y² = (x+1)² + (y+2)²i² + 2 (x+1) (y+2)i

⇒ x² + y² = x²+1 + 2x − y² − 1 + 2y + 2 (x+1) (y+2)i

⇒ 2y² − 2x + 4y + 4 = 2i (x+1) (y+2)

⇒ y² − x + 2y + 2 = i (x+1) (y+2)

On comparing both sides, we get,

⇒ (x+1) (y+2) = 0

⇒ x = −1 and y = −2

Also, y² − x + 2y + 2 = 0

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

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

Taking y = −2, (4 − x −4 + 2) = 0

⇒ x = 2

Therefore, z = x + iy = 2 − 2i.

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

Solution:

We are given,

⇒ (1+i)²ⁿ = (1−i)²ⁿ

⇒\left(\frac{1+i}{1-i}\right)^{2n} = 1

⇒\left[\frac{(1+i)^2}{(1-i)(1+i)}\right]^{2n} = 1

⇒\left[\frac{1+i^2+2i}{1-i^2}\right]^{2n} = 1

⇒\left[\frac{2i}{2}\right]^{2n} = 1

⇒ i²ⁿ = 1

⇒ i²ⁿ = i⁴

⇒ 2n = 4

⇒ n = 2

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

Question 25. If z₁, z₂, z₃ are complex numbers such that |z₁| = |z₂| = |z₃| =|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}| = 1, then find the value of |z₁ + z₂ + z₃|.

Solution:

We are given,

|z₁| = |z₂| = |z₃| =|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}| = 1

Now, |z₁ + z₂ + z₃| =|\frac{z_1\bar{z_1}}{\bar{z_1}}+\frac{z_2\bar{z_2}}{\bar{z_2}}+\frac{z_3\bar{z_3}}{\bar{z_3}}|

=|\frac{|z_1|^2}{\bar{z_1}}+\frac{|z_2|^2}{\bar{z_2}}+\frac{|z_3|^2}{\bar{z_3}}|

=|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|

= 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,

⇒ z² + |z|² = 0

⇒ (x + iy)² + |x + iy|² = 0

⇒ x² + i²y² + 2xyi + x² + y² = 0

⇒ x² − y² + 2xyi + x² + y² = 0

⇒ 2x² + 2xyi = 0

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

⇒ 2x² = 0 and 2xy = 0

⇒ x = 0 and y ∈ R

Therefore, z = 0 + iy, where y ∈ R.

Read More:

• Class 11 RD Sharma Solutions - Chapter 13 Complex Numbers - Exercise 13.1
• Class 11 RD Sharma Solutions - Chapter 13 Complex Numbers - Exercise 13.2 | Set 1