Moduli of complex numbers

Question

In $\mathbb{C}$, a set of complex numbers, there are z and w, such that $|z|=|w|$.

How can I show that: $|z+w|^2+|z-w|^2=4|z|^2$?

I'm trying to do:

$|z+w|^2+|z-w|^2=\sqrt{(z+w)^2}+\sqrt{(z-w)^2}$

The problem is that solving the square of the binomial I cannot cut the elements inside the roots, because I cannot join them in a single root. How can I demonstrate that expression?

Answer 1

Recall that

$${z\bar{z}=|z|^2\ \forall\ z \in \mathbb{C}}$$

So

$${|z+w|^2 = (z+w)(\overline{z+w})=(z+w)(\bar{z}+\bar{w})}$$

If you expand out and reuse that fact again

$${\Rightarrow z\bar{z}+z\bar{w}+\bar{z}w+w\bar{w}=|z|^2 + |w|^2 + z\bar{w} + \bar{z}w}$$

I won't write all the working once again, but similarly for ${|z-w|^2}$ you end up with

$${|z|^2 + |w|^2 - \bar{z}w - z\bar{w}}$$

And so adding both together, we have

$${\Rightarrow |z+w|^2 + |z-w|^2 = 2|z|^2 + 2|w|^2}$$

And since ${|w| = |z|}$,

$${=4|z|^2}$$

Note also that the expression in your question:

$${|z+w|^2 = \sqrt{(z+w)^2}}$$

is incorrect. We actually have

$${|z+w|^2 = \left(\Re(z) + \Re(w)\right)^2 + \left(\Im(z) + \Im(w)\right)^2}$$