Proof that $\Re\left[\frac{w+z}{w-z}\right]=\frac{|w|^2-|z|^2}{|w-z|^2}$

Question

$w$ and $z$ are given in polar form as $w=Re^{i\Phi}$, $z=re^{i\theta}$, $R>r>=0$;

I can prove this using decomposition into $w=R(\cos\Phi+i\sin\Phi)$ and then use trig identities but I want a simpler way to see why this is the case. I know the real part of a complex quotient in standard form is $\Re\left[\frac{a+bi}{c+di}\right]=\frac{ac+bd}{c^2+d^2}$. I also know that $\Re(z^2)=|z|^2$. But I don't know exactly how to express the result in the title in terms as plain as these. One idea that I have is that using difference of two squares $\frac{w+z}{w-z}=\frac{w^2-z^2}{(w-z)^2}$, which is much closer to the desired RHS. But I can't find the next step.

Answer 1

$$\frac{w+z}{w-z}=\frac{(w+z)(\bar{w}-\bar{z})}{(w-z)(\bar{w}-\bar{z})}$$ $$=\frac{|w|^2-|z|^2-(\bar{z}w-z\bar{w})}{|w-z|^2}.$$ Here $(\bar{z}w -z \bar{w})$ is purely imaginary.

$$ Re \left(\frac{w+z}{w-z}\right)= \frac{|w|^2-|z|^2}{|w-z|^2}.$$

Answer 2

\begin{align} 2\Re\left(\frac{z+w}{z-w}\right)&=\frac{z+w}{z-w}+\overline{\left(\frac{z+w}{z-w} \right)}=\frac{z+w}{z-w}+\frac{\overline{z+w}}{\overline{z-w}}\\ &=\frac{(z+w)(\overline z-\overline w)+(z-w)(\overline z+\overline w) }{(z-w)\overline{(z-w)}}\\ &=\frac{2z\overline z-2w\overline w}{|z-w|^2} \end{align} etc.

Answer 3

Well, i you use $w=a+bi, z=c+di$ you can solve your problem operating.

$\begin{equation} \begin{aligned} \frac{w+z}{w-z} &= \frac{(a+c)+(b+d)i}{(a-c)+(b-d)i} \\ & = \frac{(a+c)+(b+d)i}{(a-c)^2+(b-d)^2}\cdot((a-c)-(b-d)i) \\ & = \frac{[(a+c)(a-c)-(b+d)(d-b)]+[(a+c)(d-b)+(b+d)(a-c)]i}{(a-c)^2+(b-d)^2} \\ & = \frac{[(a^2-c^2)+(b^2-d^2)]+[(ad-ab+cd-bc)+(ab-bc+ad-cd)]i}{(a-c)^2+(b-d)^2} \\ & = \frac{[(a^2+b^2)-(c^2+d^2)]+[2(ad-bc)]i}{(a-c)^2+(b-d)^2} \\ & = \frac{[(a^2+b^2)-(c^2+d^2)]}{(a-c)^2+(b-d)^2} + \frac{[2(ad-bc)]i}{(a-c)^2+(b-d)^2}\\ & = \frac{|w|^2-|z|^2}{|w-z|^2} + \frac{[2(ad-bc)]i}{|w-z|^2} \end{aligned} \end{equation}$

then $\Re\left[\frac{w+z}{w-z}\right]=\frac{|w|^2-|z|^2}{|w-z|^2}$