Automorphisms of the unit disc is less than 1

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SUMMARY

The modulus of the automorphism defined by the formula (a-z)/(1-̄az) is strictly bounded by 1 within the unit disc. The proof utilizes the Schwarz lemma to establish that |a-z| ≤ |1 - ̄az| for all |z| ≤ 1. By manipulating the inequality and performing multiplication, it is shown that (1-|z|²)(1-|a|²) > 0, confirming the boundedness of the automorphism.

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  • Understanding of complex analysis concepts, particularly automorphisms of the unit disc.
  • Familiarity with the Schwarz lemma and its applications.
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Likemath2014
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I want to show that the modulus of the automorphism

[tex]\frac{a-z}{1-\overline{a}z}[/tex]

is strictly bounded by 1 in the unit disc. Applying Schwarz lemma gives the result immediately. But I am looking for a straight forward proof for that.

Thanks in advance
 
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So you need to prove that ##|a-z| \leq |1 - \overline{a}z|## whenever ##|z| \leq 1##. Equivalently, you require
$$(a-z)(\overline{a}-\overline{z}) \leq (1 - \overline{a}z)(1 - a\overline{z})$$
Performing the multiplication on both sides, we need
$$|a|^2 - 2\text{Re}(\overline{a}z) + |z|^2 \leq 1 - 2\text{Re}(\overline{a}z) + |a|^2|z|^2$$
It should be straightforward from this point.
 
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Now it is clear, and the last one is true because
[tex](1-|z|^2)(1-|a|^2)>0.[/tex]
Thanks
 

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