Why can we assume that $z \in \mathbb{R}$? I am looking at the solution to the following question:
Let $z, w \in \mathbb{C}, |z|, |w| <  1$. Show that:
$$
\left| \frac{z - w}{1 - z\overline{w}} \right| < 1
$$
The solution starts by assuming w.l.o.g that $z \in \mathbb{R}$ (from there it's pretty simple manipulation). 
Why can we assume that? I guess that for $z \in \mathbb{R}$ we get a maximum for this expression, but I'm not sure how to show that.
Thanks!
 A: Let $z = r e^{i \theta}$. Then, dividing the numerator by $e^{i \theta}$ makes no difference to the absolute value
$$
\left| \frac{r e^{i \theta} - w}{ 1 - r e^{i \theta} \overline{w}} \right| = \left| \frac{r - w e^{- i \theta }}{1 - r ( \overline{w e^{- i \theta}} )} \right |.
$$
and $| z | = | r | < 1$. In addition, $| w | = | w e^{- i \theta} | < 1$ too. I hope that answers your question!
A: We can assume that $z\in\mathbb{R}$ since multiplying the fraction by a complex number of modulus $1$ doesn't change its modulus, and the inequality is generic in $w$. If we have $z = re^{i\varphi}$, then  we can see
$$\begin{align}
\frac{z-w}{1-z\overline{w}} &= \frac{re^{i\varphi} - w}{1-re^{i\varphi}\overline{w}}\\
&= e^{i\varphi} \frac{r - (e^{-i\varphi}w)}{1-r\overline{e^{-i\varphi}w}},
\end{align}$$
so
$$\left\lvert \frac{z-w}{1-z\overline{w}}\right\rvert = \left\lvert\frac{r - (e^{-i\varphi}w)}{1-r\overline{e^{-i\varphi}w}}\right\rvert.$$
I prefer a proof without such assumptions, we have
$$\begin{align}
\left\lvert 1- z\overline{w}\right\rvert^2 - \lvert z-w\rvert^2
&= 1 - z\overline{w} - \overline{z} w + \lvert z\rvert^2\lvert w\rvert^2 - (\lvert z\rvert^2 - z\overline{w} - \overline{z}w + \lvert w\rvert^2)\\
&= 1 - \lvert z\rvert^2 - \lvert w\rvert^2 + \lvert z\rvert^2\lvert w\rvert^2\\
&= (1-\lvert z\rvert^2)(1-\lvert w\rvert^2),
\end{align}$$
and we can see that $\lvert 1-z\overline{w}\rvert > \lvert z-w\rvert$ for $z,w$ in the unit disk.
A: Say $z=r\times e^{ia}$
Then $$
\left| \frac{z - w}{1 - z\overline{w}} \right| =
\left| \frac{r - (e^{-ia}w)}{1 - r\overline{e^{-ia}w}} \right| 
$$
Now replace $w$ with $we^{-ia}$. They have same magnitude.
