<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] Yet another approach, is to note that $Re(\overline z_1z_2)=z_1 \cdot z_2$, where $\cdot$ denotes the euclidean dot product in $R^2$. 
Let $z=x+iy$ for $x,y$ in $R$, then we have $Re(\frac{1-z}{1+z})=\overline{1-z}{\frac{1}{1+z}}=(1-x,y)\cdot (\frac{1+x}{1+2x+x^2+y^2},\frac{-y}{1+2x+x^2+y^2})=\frac{1-(x^2+y^2)}{2+2x}=0$  
</s>