Problem 1605 (difficulty: 7/10)

Let \(\displaystyle a,b \in \C\) and \(\displaystyle |b|<1\). Prove that

\(\displaystyle \frac1{2\pi}\int_{|z|=1}\left|\frac{z-a}{z-b}\right|^2|\dz| = \frac{|a-b|^2}{1-|b|^2}+1. \)

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