Why can we assume that $z \in \mathbb{R}$?

Question

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!

Maybe analytic continuation? I mean maybe it's a trick that I don't see but usually my first guess would be that. Although I don't see how. — Patrick Da Silva, Feb 22 '14 at 13:02
http://math.stackexchange.com/questions/674667/properties-of-complex-modulus-question — lab bhattacharjee, Feb 22 '14 at 14:19

score 5 · Accepted Answer · answered Feb 22 '14 at 13:05

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!

score 4 · Answer 2 · answered Feb 22 '14 at 13:07

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.

score 1 · Answer 3 · answered Feb 22 '14 at 13:05

1

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.

answered Feb 22 '14 at 13:05

Gabriel Romon

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Why can we assume that $z \in \mathbb{R}$?

3 Answers3