Problem with absolute value of addition of complex numbers $$|z_1+z_2|^2+|z_1-z_2|^2=2|z_1|^2+2|z_2|^2$$
We need to prove that the equation is true for all $z_1$ and $z_2$ complex numbers.
How can we solve this? What I really don't understand is how to take an absolute value of addition of complex numbers (for just one complex number, for example $z$, it's $|z|=\sqrt{a^2+b^2}$) .
 A: Hint: use $$|z|^2 = z\cdot \overline{z}$$
Then $$|z+w|^2 = (z+w)(\overline{z}+\overline{w})=...$$

Or you can remember than we have parallelogram identity:
$$ e^2+f^2 = 2a^2+2b^2$$ where $a,b$ are sides in parallelogram and $e,f$ are diagonals.
A: 
What I really don't understand is how to take an absolute value of addition of complex numbers 

If $z_1 = a + b i$ and $z_2 = c+di$ then $z_1 + z_2 = (a + bi) + (c+di) = (a+c) + (b+d)i$ so $|z_1 + z_2| = \sqrt {(a+b)^2 + (c +d)^2}$.
So $|z_1 + z_2|^2 + |z_1 - z_2|^2 = \sqrt{(a+b)^2 + (c+d)^2}^2 + \sqrt{(a-b)^2 + (c-d)^2}^2 =(a+b)^2 + (c+d)^2 + (a-b)^2 + (c-d)^2$ which should be very easy to finish.
....
Another way of thinking of this is noting $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$, and $|z|^2 = z\overline{z}$ 
So $|z_1 + z_2|^2 = (z_1 + z_2)(\overline{z_1} + \overline{z_2}) = z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1} + z_2\overline{z_2} = |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + z_2\overline{z_1}$ 
and it's just a matter of showing $z_1\overline{z_2} = - z_2\overline{z_1}$.
i.e. that $(Re(z_1) + Im(z_1)i)(Re(z_2) - Im(z_2)i) = -(Re(z_1) - Im(z_1)i)(Re(z_2) + Im(z_2)i)$
