Question

If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = `pi/2`, then show that `barz`w = –i.

Answer 1

Let z = r₁ (cosθ₁ + isinθ₁) and w = r₂ (cosθ₂ + isinθ₂)

zw = r₁r₂ [(cosθ₁ + isinθ₁)] [(cosθ₂ + isinθ₂)]

|zw| = r₁r₂ = 1

Now arg(z) – arg(w) = `pi/2`

θ₁ – θ₂ = `pi/2`

⇒ arg `(z/w) = pi/2`

`barzw` = r₁ (cosθ₁ – isinθ₁) r₂ (cosθ₂ + isinθ₂)

= r₁ r₂ [cosθ₁ cosθ₂ + icosθ₁ sinθ₂ – isinθ₁ cosθ₂ – i₂ sinθ₁ sinθ₂]

= r₁ r₂ [(cosθ₁ cosθ₂ + sinθ₁ sinθ₂) + i(cosθ₁ sinθ₂ – sinθ₁ cosθ₂)]

= r₁ r₂ [cos(θ₂ – θ₁) + isin(θ₂ – θ₁)]

= `r_1r_2 [cos((-pi)/2) + i sin((-pi)/2)]`

= `r_1r_2 [cos  pi/2 - i sin  pi/2]`

= 1  .....[0 – i]

Here `barzw` = –i.

Hence proved.