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Theorem:

Let $w, z \in \mathbb{C}$ such that $\bar{z}w \neq 1$. Then $$ |z| < 1 \land |w| < 1 \Rightarrow \left| \frac{w-z}{1-\bar{w}z}\right| < 1$$

I tried to make $w = u +iv, z = x + iy$ and expand, but that did not give me useful results. How to prove it?

Community
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Henricus V.
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1 Answers1

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Equivalently $$|w-z|^2<|1-\bar{w}z|^2$$ i.e. $$(w-z)(\bar{w}-\bar{z})<(1-\bar{w}z)(1-w\bar{z})$$ which is also equivalent to $$|w|^2+|z|^2<1+|w|^2|z|^2$$ The above inequality can be written as $$|w|^2(1-|z|^2)<1-|z|^2$$ which is always true for $|w|<1$, $|z|<1$.

RTJ
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  • On which step does it imply the condition for $|z|<1$ cannot be lifted? – Henricus V. Dec 06 '15 at 19:40
  • @HenryW The last step. You need $1-|z|^2>0$ when you multiply both sides of $|w|^2<1$ so that the inequality direction remains unchanged. – RTJ Dec 06 '15 at 19:44