Let $z$ and $w$ be complex numbers. Proof that ${\rm Re}\left( \dfrac{w+z}{w-z} \right)= \dfrac{|w|^2-|z|^2}{|w-z|^2}$
I defined my complex numbers as $z=a+bi$ and $w=c+di$.
Now I got that the real part of $\dfrac{w+z}{w-z}=\dfrac{a+c}{a-c}$
Then I used the property $|z|^2=z \cdot z^*$ being $z^*$ the conjugate of $z$. But I did not get something clear. How would do you proceed?
$$ \frac{w+z}{w-z} =\frac{w+z}{w-z}\cdot\frac{w^*-z^*}{w^*-z^*} =\frac{ww^*+zw^*-wz^*-zz^*}{(w-z)(w-z)^*} =\frac{|w|^2-|z|^2}{|w-z|^2}+\frac{zw^*-(zw^*)^*}{|w-z|^2}. $$