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I am struggling to show that, for any $\alpha, z \in \mathbb C$ such that $|\alpha|<1$ and $|z|<1$:

$$\left | \frac{z - \alpha}{1-\bar\alpha z} \right | < 1$$

This should be doable with simple algebric manipulations, without using the properties of Mobius transformations, but I cannot figure out how.

Thank you for your time.

Riccardo Orlando
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    This has been asked many times so it will be marked as duplicate - eliminating denominators,squaring and noting that the real part in the modulus square is the same, it reduces to $(1-|z|^2)(1-|\alpha|^2)>0$ – Conrad Sep 03 '19 at 10:41
  • I found the answer that says it reduces to that, but I couldn't find the explicit step-by-step solution anywhere. – Riccardo Orlando Sep 03 '19 at 10:45
  • $|z-\alpha|^2=|z|^2+|\alpha|^2 -2\Re{\bar \alpha z}$ and $|1-\bar \alpha z|^=1+|\alpha |^2|z|^2--2\Re{\bar \alpha z}$ – Conrad Sep 03 '19 at 10:57

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