z1 and z2 are two complex numbers such

According to the question,

Let z1 = |z1| (cos θ1 + I sin θ1) and z2 = |z2| (cos θ2 + I sin θ2)

Given that |z1| = |z2|

And arg (z1) + arg (z2) = π

⇒ θ1 + θ2 = π

⇒ θ1 = π – θ2

Now, z1 = |z2| (cos (π – θ2) + I sin (π – θ2))

⇒ z1 = |z2| (-cos θ2 + I sin θ2)

⇒ z1 = -|z2| (cos θ2 – I sin θ2)

⇒ z1 = – [|z2| (cos θ2 – I sin θ2)]

Hence, z1 = -z̅2

Hence proved.