How do I prove this parallelogram equality? |z1+z2|^2 +|z1-z2|^2 = 2(|z1|^2+|z2|^2)

Question

anonymous · Answer

Is z a complex number ?

anonymous · Answer

Yes

anonymous · Answer

I just don't know what to do with the absolute if I regard it as positive only I get:

anonymous · Answer

$$z_1^2+2z_1z_2+z_2^2+z_1^2-2z_1z_2+z_2^2 $$

anonymous · Answer

$$2(z_1^2+z_2^2)$$

anonymous · Answer

but then I don't have the absolute signs

anonymous · Answer

First, if z is a complex number we have :
$$|z|^2=z\bar z$$
So :
$$RHS=|z_1+z_2|^2+|z_1-z_2|^2\
~~~~~~=(z_1+z_2)(\overline{z_1+z_2})+(z_1-z_2)(\overline{z_1-z_2})\
~~~~~~=z_1\overline {z_1}+z_1\overline {z_2}+z_2\overline {z_1}+z_2\overline {z_2}+z_1\overline {z_1}-z_1\overline {z_2}-z_2\overline {z_1}+z_2\overline {z_2}\
~~~~~=2z_1\overline {z_1}+2z_2\overline {z_2}\
~~~~~~=2(|z_1|^2+|z_2|^2)\
~~~~=LHS
$$

anonymous · Answer

Oh right I forgot about that! Thank you I get it now