# complex numbers - proof of this statement

i am trying to prove this statement, i dont but how to start.

$$\forall z,w \in \mathbb{C}\quad |z|^2+|w|^2=\frac{1}{2}(|z+w|^2+|z-w|^2)$$

can someone please show me how start?

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Use that $|x|^2 = x \cdot \overline{x}$ and distribute all multiplications. – WimC Dec 1 '12 at 17:11
@WimC I wanted you to post it as answer so that I can upvote it. – Amr Dec 1 '12 at 17:18

If $z=r_1 cis\theta$ and $w=r_2cis\phi$

$|z+w|^2+|z-w|^2$ $=|(r_1\cos\theta+r_2\cos\phi)+i(r_1\sin\theta+r_2\sin\phi)|^2+|(r_1\cos\theta-r_2\cos\phi)+i(r_1\sin\theta-r_2\sin\phi)|^2$ $=(r_1\cos\theta+r_2\cos\phi)^2+(r_1\sin\theta+r_2\sin\phi)^2+(r_1\cos\theta-r_2\cos\phi)^2+(r_1\sin\theta+r_2\sin\phi)^2$ $=2(r_1^2+r_2^2)=2(|z|^2+|w|^2)$

Alternatively,

If $z=x+iy$ and $w=a+ib$

$|z+w|^2+|z-w|^2$ $=|(x+a)+i(y+b)|^2+|(x-a)+i(y-b)|^2$ $=(x+a)^2+(y+b)^2+(x-a)^2+(y-b)^2$ $=2(x^2+y^2+a^2+b^2)=2(|z|^2+|w|^2)$

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thanks great explanation – doniyor Dec 1 '12 at 17:35
@doniyor, welcome. Please have a look into the edited answer? – lab bhattacharjee Dec 1 '12 at 18:17
yeah, i saw it. i am learning the properties now. thank you so much, you explained better than the book :) – doniyor Dec 1 '12 at 18:28

On request: Use $|x|^2 = x \cdot \overline{x}$ and distribute all the multiplications.

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yes. +1 for nice answer – Amr Dec 1 '12 at 17:26