Find the real values of x and y for which:\(\frac{(x+3i)}{(2+iy)}\) =

1.	Find the real values of x and y for which:\(\frac{(x+3i)}{(2+iy)}\) = (1 - i)(x+3i)/(2+iy)
Answer» Given: \(\frac{(x+3i)}{(2+iy)}\) = (1 - i) ⇒ x + 3i = (1 – i)(2 + iy) ⇒ x + 3i = 1(2 + iy) – i(2 + iy) ⇒ x + 3i = 2 + iy – 2i – i 2y ⇒ x + 3i = 2 + i(y – 2) – (-1)y [i² = -1] ⇒ x + 3i = 2 + i(y – 2) + y ⇒ x + 3i = (2 + y) + i(y – 2) Comparing the real parts, we get x = 2 + y ⇒ x – y = 2 …(i) Comparing the imaginary parts, we get 3 = y – 2 ⇒ y = 3 + 2 ⇒ y = 5 Putting the value of y = 5 in eq. (i), we get x – 5 = 2 ⇒ x = 2 + 5 ⇒ x = 7 Hence, the value of x = 7 and y = 5

Find the real values of x and y for which:\(\frac{(x+3i)}{(2+iy)}\) = (1 - i)(x+3i)/(2+iy)

Answer»

Given: \(\frac{(x+3i)}{(2+iy)}\) = (1 - i)

⇒ x + 3i = (1 – i)(2 + iy)

⇒ x + 3i = 1(2 + iy) – i(2 + iy)

⇒ x + 3i = 2 + iy – 2i – i 2y

⇒ x + 3i = 2 + i(y – 2) – (-1)y [i² = -1]

⇒ x + 3i = 2 + i(y – 2) + y

⇒ x + 3i = (2 + y) + i(y – 2)

Comparing the real parts, we get

x = 2 + y

⇒ x – y = 2 …(i)

Comparing the imaginary parts, we get

3 = y – 2

⇒ y = 3 + 2

⇒ y = 5

Putting the value of y = 5 in eq. (i), we get

x – 5 = 2

⇒ x = 2 + 5

⇒ x = 7

Hence, the value of x = 7 and y = 5