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Let $z$ and $w$ be complex numbers with $|z|< 1, |w|<1$.

Show that $|1-\overline zw|^2-|z-w|^2=(1-|z|^2)(1-|w|^2)$, and hence prove that $|\frac{z-w}{1-\overline zw}|<1 $

I'm having trouble getting started with the first part. I tried expanding the numbers out with their real/imaginary components but got a giant mess that I couldn't work with. Could someone point me in the right direction?

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Hint $$|1-\overline zw|^2-|z-w|^2 =(1-\overline{z}w)(1-z\overline{w})-(z-w)(\overline{z}-\overline{w}).$$

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  • $\begingroup$ Thanks, managed to figure it out with that hint but I'm still having trouble grasping how those two equations you stated are equal, which properties did you use to get that? $\endgroup$ – Forcefedglas Mar 2 '16 at 11:12
  • $\begingroup$ In general $|z|^2 = z\overline{z}$. $\endgroup$ – C. Dubussy Mar 2 '16 at 11:18

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