proof $ \frac{z-z'\bar{z}}{1-z'} \in \Bbb R $ is real.

Question

How to prove that $$ \frac{z-z'\bar{z}}{1-z'} \in \Bbb R $$ given that $z', z \in \Bbb{C}$ , and $|z'|=1$.

Answer 1

For a nonzero complex number $w$ we have $\frac{1}{w} = \frac{\bar{w}}{|w|^2}$. Taking $w=1-z'$, we have

$$\frac{z-z'\bar{z}}{1-z'} = \frac{(z-z'\bar{z})\overline{(1-z')}}{|1-z'|^2}.$$

Using $|z'|^2=1$, the numerator can be expanded as $z - z'\bar{z} - \overline{z'} z + \bar{z}$. Check that the imaginary part of this quantity is zero.

Answer 2

A complex number is real iff it's the same as its conjugate.

From $1=1^2=|z'|^2=z'\bar z'$ we know $\bar z'=1/z'$. Now $$\overline{\frac{z-z'\bar{z}}{1-z'}}=\frac{\bar z-\bar z'\bar{\bar z}}{1-\bar z'}=\frac{\bar z-1/z'\cdot z}{1- 1/z'}=\frac{z'\bar z- z}{z'-1}=\frac{z-z'\bar{z}}{1-z'}.$$