<s>[INST] Complex Expression is Real 
If I have the a complex number $z \in \mathbb{C}$ with absolute value $|z| = 1$, how do I show that $-i \frac {z-1}{z+1}$ is real? 

 [/INST] A number $w$ is real iff $\operatorname{Im}(w) = \frac{1}{2i}(w - \bar w) = 0\,$. In this case:
$$\require{cancel}
\begin{align}
-i \frac {z-1}{z+1} - \overline{\left(-i \frac {z-1}{z+1}\right)} &= -i \frac {z-1}{z+1} - \left(+i \frac{\bar z - 1}{\bar z + 1}\right) \\
 &= -i \cdot\frac{(z-1)(\bar z + 1)+ (\bar z -1)(z+1)}{(z+1)(\bar z + 1)} \\
 &= -i \frac{|z|^2+\bcancel{z}-\cancel{\bar z}-1+|z|^2+\cancel{\bar z} - \bcancel{z} - 1}{|z+1|^2} = \\
 &= -2i \frac{\cancel{|z|^2}-\cancel{1}}{|z+1|^2} \\
 &= 0
\end{align}
$$
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