How Does the Triangle Inequality Apply to Complex Numbers?

[Poirot1]

let z,w be complex numbers. Prove:

2|z||w| <_ |z|^2 + |w|^2

 

[Fantini]

Well, you could note that $(|z| - |w|)^2 \geq 0$ and work from there.

 

[Poirot1]

0<_ (|z|-|w|)^2=|z|^2+|w|^2-2Re(zw*).

Ok so I can say Re(zw*)<_|zw*|=|zw| but that doesn't quite get me there.

 

[Fantini]

They are both real numbers, so when you expand $(|z| - |w|)^2$ you get $|z|^2 - 2|z| |w| + |w|^2 \geq 0$, hence the result.

There is the alternative of using that

$$|z|^2 + |w|^2 = \Re^2 (z) + \Im^2 (z) + \Re^2 (w) + \Im^2(w)$$

and

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

Now, $a = \Re^2 (z) + \Im^2 (z)$ is a positive real number, same as $b = \Re^2 (w) + \Im^2 (w)$. This is the arithmetic geometric mean written in a different way:

$$\frac{a+b}{2} \geq \sqrt{ab} \implies \frac{\Re^2 (z) + \Im^2 (z) + \Re^2 (w) + \Im^2(w)}{2} \geq \sqrt{(\Re^2 (z) + \Im^2 (z))(\Re^2 (w) + \Im^2 (w))}.$$

How did you get that $(|z| - |w|)^2 = |z|^2 + |w|^2 - 2 \Re (z \bar{w})$?