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?
 A: $$
a^2+b^2=|a+ib|^2=\left|\frac{x-iy}{x+iy}\right|^2
=\left|\frac{re^{-i\theta}}{re^{i\theta}}\right|^2=1
$$
where $re^{i\theta}$ is simply $x+iy$ written in its polar form.
A: 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
$$ a^2+b^2 = \frac{(x^2-y^2)^2+(2xy)^2}{(x^2+y^2)^2}=\frac{x^4 - 2x^2y^2+y^4 + 4x^2y^2)}{x^4 +2x^2y^2+y^4} = \frac{x^4 +2x^2y^2+y^4}{x^4 +2x^2y^2+y^4} = 1
$$
A: 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:
$$a^2 + b^2 = |a + bi|^2 = \left| \frac{x - yi}{x + yi} \right|^2 = \frac{|x-yi|^2}{|x + yi|^2} = \frac{x^2 + y^2}{x^2 + y^2} = 1$$
A: ${x-yi\over{x+yi}}= {x-yi\over{x+yi}}{x-yi\over{x-yi}} ={x^2 - y^2 - 2xyi \over{x^2 + y^2}}={x^2 - y^2\over x^2 + y^2} + {-2xy\over x^2 + y^2}i$
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 $a^2 + b^2 = ({1\over x^2 + y^2})^2(x^4 + 2x^2y^2 + y^4)=({1\over x^2 + y^2})^2(x^2 + y^2)^2 = 1$
