# z1 and z2 are two complex numbers such that |z1| = |z2| and arg (z1) + arg (z2) = π, then show that 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)]

z1 = -z̅ 2

Hence proved.

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