Simplification of $|a+b|^2$ for $a,b \in \mathbb{C}$

Question

How do I simplify $|a+b|^2$, where $a,b \in \mathbb{C}$ and $|a|=|b|=1$? I know that the result is $4-|a-b|^2$, but I would like to be be explained how to do the simplification in the most elegant way.

Answer 1

It is

$$|a+b|^2=(a+b)\overline{(a+b)}=(a+b)(\overline{a}+\overline{b})=a\overline{a}+b\overline{b}+a\overline{b}+b\overline{a}=2+a\overline{b}+b\overline{a}$$

$$|a-b|^2=(a-b)\overline{(a-b)}=(a-b)(\overline{a}-\overline{b})=a\overline{a}+b\overline{b}-a\overline{b}-b\overline{a}=2-a\overline{b}-b\overline{a}$$

Now, if you add both equalities you get

$$|a+b|^2+|a-b|^2=4.$$