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Prove:

If $|z|,|w|<1$, then $\left|\frac{z-w}{1-z\bar{w}}\right|<1$.

Not quite sure how to approach this.. I've tried squaring it and to do something from there:

$$\left|\frac{z-w}{1-z\bar{w}}\right|^2<1^2$$ or $$\frac{z-w}{1-z\bar{w}}*\overline{\frac{z-w}{1-z\bar{w}}}<1$$

I started to multiply all the $z$ and $w$ but then it appears that I reached a dead end.

Also the second part is, Prove:

If $|w|=|z|=1$, then $\frac{z-w}{1-zw}\in{\Re}$.

Don't have a clue on this one..

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    For the first part, see https://math.stackexchange.com/q/1190613/42969. – Martin R May 30 '18 at 13:13
  • Looks like my book has a mistake because in the link you just gave, the lower $w$ is a conjugent $w$, looks like that is the correct way to write the exercise. –  May 30 '18 at 13:32
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    $\left|\frac{z-w}{1-zw}\right|<1$ (as written in the question, without conjugating $w$ in the denominator) is actually wrong: try $z = i/2, w = -i/2$. – Martin R May 30 '18 at 14:01
  • Very nice, I'll edit it accordingly, thanks a lot - been breaking my head on some book's mistake XD. –  May 30 '18 at 14:06

2 Answers2

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Let $|w|=|z|=1$

$\frac{z-w}{1-wz}=\frac{\bar{z}-\bar{w}}{1-\bar{z}\bar{w}}$

But:

$(1-\bar{w}\bar{z})(z-w)=z-\bar{w}-w+\bar{z}=(\bar{z}-\bar{w})(1-zw)$

From this calculation we have that the above fractional complex number is equal to its conjugate,thus it is a real number.

Marios Gretsas
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For the second part:

You only need to show that for $r = \frac{z-w}{1-z w}$ holds $\bar r = r$.

We have $$|z| = |w|= 1 \Rightarrow \frac{1}{z} = \bar z \mbox{ and } \frac{1}{w} = \bar w$$

Now,

$$\bar r = \frac{\bar z - \bar w}{1-\bar z \bar w} = \frac{\frac{1}{z} - \frac{1}{w}}{1-\frac{1}{zw}} = \frac{w - z}{wz-1} = r$$ So, second part is also done.

trancelocation
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