You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.2

Question 1.

Evaluate the following if z = 5 – 2i and w = -1 + 3i

(i) z + w

(ii) z – iw

(iii) 2z + 3w

(iv) zw

(v) z^{2} + 2zw+ w^{2}

(vi) (z + w)^{2}

Solution:

(i) z + w = (5 – 2i) + (-1 + 3i)

= 4 + i

(ii) z – iw= (5 – 2i) -i (-1 + 3i)

= 5 – 2i + i + 3i² (∵ i² = -1)

= 5 – 2i + i – 3(-1)

= 5 – 2i + i + 3

= 8 – i

(iii) 2z + 3w = 2 (5 – 2i) + 3 (- 1 +3i)

= 10 – 4i – 3 + 9i

= 7 + 5i

(iv) zw = (5 – 2i) (- 1 + 3i)

= -5 + 2i + 15i – 6i²

= -5 + 17i + 6

= 1 + 17i

(v) z^{2} + 2zw + w^{2}

= (z + w)²

= (4 + i)²

= (4) + 2(4)(i) + (i)²

= 16 + 8i + i²

= 15 + 8i

(vi) (z + w)^{2 }= z² + 2zw + w²

= (z + w)² = (4 + i)²

= 16 + 8i + i²

= 15 + 8i

Question 2.

Given the complex number z = 2 + 3i, represent the complex numbers in the Argand diagram.

(i) z, iz, and z + iz

(ii) z, -iz, and z – iz

Solution:

(i) z, iz and z + iz.

z = 2 + 3i

iz = i(2 + 3i) = -3 + 2i

z + iz = 2 + 3i – 3 + 2i = -1 + 5i

(ii) z = 2 + 3i

-iz = -i(2 + 3i)

= -2i – 3i^{2}

= (3 – 2i)

z – iz = (2 + 3i) + (3 – 2i)

= 5 + i

Question 3.

Find the values of the real numbers x and y, if the complex numbers.

(3 – i) x – (2 – i) y + 2i + 5 and 2x + (-1 + 2i) y + 3 + 2i are equal

Solution:

Given that the complex numbers are equal

(3 – i) x -(2 – i) y + 3 + 2i + 5

= 2x + (-1 + 2i)y + 3 + 2i

3x – ix – 2y + iy + 2i +5

= 2x – y + 2iy + 3 + 2i

(3x – 2y + 5) + i(y – x + 2)

= (2x – y + 3) + i(2y + 2)

Equating real and imaginary parts separately

3x – 2y + 5 = 2x – y + 3

x – y = -2 ………. (1)

y – x + 2 = 2y +

-x – y = 0 ………. (2)

solving 1 and 2

y = 1

Substituting y = 1 in (1)

x – 1 = -2

x = -2 + 1 = -1

values of x and y are -1, 1

### Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.2 Additional Problems

Question 1.

Find the real values of x and y, if

(i) (3x – 7) + 2 iy = -5y + (5 + x)i

(ii) (1 – i)x + (1 + i)y = 1 – 3i

(iii) (x + iy)(2 – 3i) = 4 + i

(iv)

Solution:

(i) We have (3x – 7) + 2 iy = 5y + (5 + x)i

⇒ 3x – 7 = 5y and 2y = 5 + x

⇒ 3x + 5y = 7 and x – 2y = -5

⇒ x = -1 y = 2

(ii) We have, (1 – i) x + (1 + i)y = 1 – 3i

⇒ (x + y) + i(-x + y) = 1 – 3i

⇒ x + y = 1

and -x + y = 3 [On equating real and imaginary parts]

⇒ x = 2 and y = -1

⇒ x + y – 2 = 0 and y – x = 10

⇒ x = -4 , y = 6.

Question 2.

Find the real values of x and y for which the complex numbers -3 + ix^{2}y and x^{2} + y + 4i are conjugate of each other.

Solution:

Since -3 + ix^{2}y and x^{2} + y + 4i are complex conjugates.

∴ -3 + ix^{2}y = x^{2} + y + 4i ….. (i) and, x^{2} y = -4 …… (ii)

Question 3.

Given x = 2 – 3i and y = 4 + i

Solution:

Do it yourself