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## 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 = 5 – 2i, w = -1 + 3i

z + w = (5 – 2i) + (-1 + 3i)

= (5 – 1) + (-2i + 3i)

= 4 + i

(ii) z – iw = (5 – 2i) – i (-1 + 3i)

= 5 – 2i + i + 3

= (5 + 3) + (-2i + i)

= 8 – i

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

= 10 – 4i – 3 + 9i

= 7 + 5 i

(iv) zw = (5 – 2i) (-1 + 3i)

= -5 + 15i + 2i – 6i^{2}

= -5 + 17i + 6

= 1 + 17i

(v) z^{2} + 2zw + w^{2} = (z + w)^{2} [from (i)]

= (4 + i)^{2}

= 16 – 1 + 8i

= 15 + 8i

(vi) (z + w)^{2} = 15 + 8z [from (v)]

Question 2.

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

(i) z, iz, and z + iz

(ii) z, -iz, and z – iz

Solution:

(i) z = 2 + 3i

iz = i(2 + 3i)

= (2i – 3)

= -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:

(3 – i) x – (2 – i) y + 2i + 5 = 2x + (-1 + 2i) y + 3 + 2i

⇒ 3x – ix – 2y + iy + 2i + 5 = 2x – y + 2yi + 3 + 2i

⇒ (3x – 2y + 5) + 1 (-x + y + 2) = (2x – y + 3) + i (2y + 2)

Equate real parts on both sides

3x – 2y + 5 = 2x – y + 3

x – y = -2 ……. (1)

Equate imaginary parts on both sides

-x + y + 2 = 2y + 2

-x – y = 0

x + y = 0 ……. (2)

(1) + (2) ⇒ 2x = -2

x = -1

Substituting x = -1 in (2)

-1 + y = 0

⇒ y = 1

∴ x = -1, y = 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

… (0 and, x^{2} y = -4 …. (ii)

Question 3.

Given x = 2 – 3i and y = 4 + i

Solution:

Do it yourself