Complex number ( prove ) Let
$$
{x-yi\over{x+yi}}=a+bi\;\;.
$$
Prove $a^2+b^2=1$
I don't know how to start prove it, can anyone help me?

Another method:
Note that $a^2 + b^2 = (a + bi)(a - bi)$, as indeed for any complex number $w = a + bi$,
$$ww^* = |w|^2 = a^2 + b^2$$
where $^*$ is the complex conjugate.
In this case $$(a + bi)^* = \left(  \frac{x - yi}{x + yi} \right)^* =  \frac{(x - yi)^*}{(x + yi)^*}  = \frac{x + yi}{x - yi}$$
Hence
$$a^2 + b^2 = (a + bi)(a + bi)^* = \frac{x - yi}{x + yi}\frac{x + yi}{x - yi}  = 1$$

And another:
$$a^2 + b^2 = |a + bi|^2 = \left| \frac{x - yi}{x + yi} \right|^2 = \frac{|x-yi|^2}{|x + yi|^2} = \frac{x^2 + y^2}{x^2 + y^2} = 1$$
