So I've been trying to solve this one for a few hours and am now out of ideas on how to approach this problem.

Here are the inequalities:

$$\text{show that if}$$

$$z,w \in \Bbb C$$ $$|z| < 1 \text{ and } |w| < 1 $$


$$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\left|\frac{z-w}{1-\overline{z}w}\right|< 1$$

My thoughts have been to first

$$\left|z-w\right|< \left| 1-\overline{z}w \right|$$

and then

$$\left|z-w\right|^2< \left| 1-\overline{z}w \right|^2 \ \ => \ \ (z-w)(\overline{z}-\overline{w})< (1-\overline{z}w)(1-z\overline{w})$$

which would give me

$$|z|^2 +|w|^2<1+|z|^2|w|^2 \ \ \ => \ \ (1-|z|^2)(1-|w|^2)>0$$

But after that I don't know where to go next, or even if I'm on the right track..

I would really appreciate some help on how to approach and solve this problem!

  • 1
    $\begingroup$ I don't know what's so special about this inequality, but this has been asked a few times before. $\endgroup$ – Git Gud Sep 6 '14 at 11:26
  • 1
    $\begingroup$ Isn't that good enough since we know $(1-|z|^2)$ and $(1-|w|^2)$ are indeed both positive since you said earlier that $|z|$ and $|w|$ are less than one? $\endgroup$ – Sheheryar Zaidi Sep 6 '14 at 11:26
  • $\begingroup$ You've done it... $\endgroup$ – Shakespeare Sep 6 '14 at 11:27
  • $\begingroup$ Oh wow... I've been too invested in this problem to notice that I had the answer.. I guess that happens after a few hours of pulling my hair, haha! Thank you! $\endgroup$ – Loc Sep 6 '14 at 11:37

Fix $w$ and consider the function $f(z)=\frac{z-w}{1-\bar{z}w}$. Now show that $f(f(z))=z$. What kind of functions satisfy the functional equation $f(f(x))=x$?


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