Convert the following in the polar form:
(i) (1 + 7i)/(2 - i)2
(ii) (1 + 3i)/(1 - 2i)
(i) Here, z = 1 + 7i/(2 - i)2
= 1 + 7i/3 - 4i
= (1 + 7i)(3 + 4i)/32 + 42
= 3 + 4i + 21i + 28i2/25
= 3 + 25i - 28/25 = -25 + 25i/25
= -1 + i
Let r cos θ = -1 and r sin θ = 1
r2(cos2θ + sin2θ) = 1 + 1 = 2
⇒ r = √2
⇒ cos θ = -1/√2, sin θ = 1/√2
⇒ θ = π - π/4 = 3π/4 (since θ lies in II quadrant)
Therefore, z = √2 (cos 3π/4 + i sin 3π/4)
(ii) Here, z = 1 + 3i/1 - 2i
= 1 + 3i/1 - 2i × 1 + 2i/1 + 2i
= 1 + 2i + 3i - 6/1 + 4
= -5 + 5i/5 = -1 + i
Let r cos θ = -1 and r sin θ = 1
r2(cos2θ + sin2θ) = 1 + 1 = 2
⇒ r = √2
⇒ cos θ = -1/√2, sin θ = 1/√2
⇒ θ = π - π/4 = 3π/4 (since θ lies in II quadrant)
Therefore, z = √2 (cos 3π/4 + i sin 3π/4)