<s>[INST] 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?
 [/INST] First off, the problem posed is flawed, because it forgot to specify that $x$ and $y$ are real.  If not, try $x=i$ and $y=1$ to see that the relation does not hold.
Now you may want to prove this without knowledge of deMoivre's theorem (complex exponentials).  Here is how:
$$\frac{x+iy}{x-iy}=\frac{x+iy}{x-iy}\frac{x+iy}{x+iy} = \frac{(x+iy)^2}{(x-iy)(x+iy)} = \frac{x^2 -y^2 + 2ixy}{x^2+y^2}$$ so $$ a = \frac{x^2 -y^2 }{x^2+y^2}\\ b = \frac{2xy}{x^2+y^2}
$$
Then
$$ a^2+b^2 = \frac{(x^2-y^2)^2+(2xy)^2}{(x^2+y^2)^2}=\frac{x^4 - 2x^2y^2+y^4 + 4x^2y^2)}{x^4 +2x^2y^2+y^4} = \frac{x^4 +2x^2y^2+y^4}{x^4 +2x^2y^2+y^4} = 1
$$
</s>