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Prove that |z1 ± z2|^2 = |z1|^2+ |z2|^2 ⇔ arg(z1)- arg(z2) = π/2

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Prove that |z1 ± z2|2 = |z1|2+ |z2|2 ⇔ arg(z1)- arg(z2) = π/2 

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⇔ θ1 - θ2 = π/2

⇔arg(z1) - arg(z2) = π/2

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