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?

${x-yi\over{x+yi}}= {x-yi\over{x+yi}}{x-yi\over{x-yi}} ={x^2 - y^2 - 2xyi \over{x^2 + y^2}}={x^2 - y^2\over x^2 + y^2} + {-2xy\over x^2 + y^2}i$
So $a= {x^2 - y^2\over x^2 + y^2}$ and $b =  {-2xy\over x^2 + y^2}$
So $a^2 = ({1\over x^2 + y^2})^2(x^4 - 2x^2y^2 + y^4)$ and $b^2 =  ({1\over x^2 + y^2})^2(4x^2y^2)$
and $a^2 + b^2 = ({1\over x^2 + y^2})^2(x^4 + 2x^2y^2 + y^4)=({1\over x^2 + y^2})^2(x^2 + y^2)^2 = 1$
