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Consider $ B_{a}\left(z\right)=\frac{z-a}{1-\overline{a}z} $

For $|a|<1$.

How can I show that $|B_a(z)|< 1$ for any $ |z|<1 $ ? Here's what I have tried:

$$ |B_{a}|^{2}=\frac{z-a}{1-\overline{a}z}\cdot\frac{\overline{z}-\overline{a}}{1-a\overline{z}}=\frac{|z|^{2}-\overline{a}z-a\overline{z}+|a|^{2}}{1-a\overline{z}-\overline{a}z+|a|^{2}|z|^{2}}=\frac{|z|^{2}-2\text{Re}\left(\overline{a}z\right)+|a|^{2}}{1-2\text{Re}\left(\overline{a}z\right)+|a|^{2}|z|^{2}} $$ But it dosent lead me anywhere so I'll be glad to see a good idea to prove it. Thanks in advance

DirichletIsaPartyPooper
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1 Answers1

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Think about it as a function of $ a$. for $|a|=1 $ and fixed $|z|<1$, you get $|B_a(z)|=1$.So by the maximum principle, since this function is not constant it does not accepts maximum points in the open disk, so for any $|a|<1 $, $|B_a(z)|<1$

FreeZe
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