<s>[INST] For complex $z$, $|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right)  = 0$ 
If $|z|=1$, show that: $$\mathrm{Re}\left(\frac{1 - z}{1 + z}\right) = 0$$

I reasoned that for $z = x + iy$, $\sqrt{x^2 + y^2} = 1\implies x^2 + y ^2 = 1$ and figured the real part would be:
$$\frac{1 - x}{1 + x}$$
I tried a number of manipulations of the equation but couldn't seem to arrive at any point where I could link the two to show that the real part was = 0.
 [/INST] Let $z=a+bi$, where $a^2+b^2=1$ and $z\neq -1$. Then $$\begin{eqnarray}
\Re\left(\frac{1-z}{1+z}\right)
&=&\Re\left(\frac{(1-z)\overline{(1+z)}}{|1+z|^2}\right)\\
&=&\Re\left(\frac{(1-a-bi)(1+a-bi)}{(a+1)^2+b^2}\right)\\
&=&\Re\left(\frac{1-(a^2+b^2)-2bi}{(a+1)^2+b^2}\right)\\
&=&\Re\left(\frac{-2bi}{(a+1)^2+b^2}\right)=0\end{eqnarray}$$
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