Question

If for complex numbers z₁ and z₂, arg (z₁) – arg (z₂) = 0, then show that `|z_1 - z_2| = |z_1| - |z_2|`.

Answer 1

Given that for z₁ and z₂, arg (z₁) – arg (z₂) = 0.

Let us represent z₁ and z₂ in polar form.

z₁ = r₁(cosθ₁ + isin θ₁) and z₂ = r₂(cosθ₂ + isin θ₂)

arg(z₁) = θ₁ and arg(z₂) = θ₂

Since arg(z₁) – arg(z₂) = 0

⇒ θ₁ – θ₂ = 0

⇒ θ₁ = θ₂ 

Now z₁ – z₂ = r₁(cosθ₁ + isinθ₁) – r₂(cosθ₂ + isin θ₂)

= r₁cosθ₁ + ir₁sinθ₁ – r₂cosθ₁ – ir₂sin θ₁   .....`[because theta_1 = theta_2]`

= (r₁cosθ₁ – r₂cosθ₁) + i(r₁sinθ₁ – r₂sinθ₁)

∴ |z₁ – z₂| = `sqrt((r_1 costheta_1 - r_2 cos theta_1)^2 + (r_1 sintheta_1 - r_2 sin theta_1)^2`

= `sqrt((r_1^2 cos^2 theta_1 + r_2^2 cos^2 theta_1 - 2r_1r_2 cos^2 theta_1 + r_1^2 sin^2 theta_1 + r_2^2 sin^2 theta_1 - 2r_1r_2 sin^2 theta_1))`

= `sqrt(r_1^2 (cos^2 theta_1 + sin^2 theta_1) + r_2^2 (cos^2 theta_1 + sin^2 theta_1) - 2r_1 r_2 (cos^2 theta_1 + sin^2 theta_1))`

= `sqrt(r_1^2 + r_2^2 - 2r_1r_2)`

= `sqrt((r_1 - r_2)^2`

= r₁ – r₂

= |z₁| – |z₂|

Hence, |z₁| – |z₂| = |z₁| – |z₂|.