Problem 1605 (difficulty: 7/10)

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

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

Solution:

\(\displaystyle \frac1{2\pi}\int_{|z|=1}\left|\frac{z-a}{z-b}\right|^2|\dz| = \frac1{2\pi}\int_{|z|=1} \frac{(z-a)(\overline{z}-\overline{a})}{(z-b)(\overline{z}-\overline{b})} \cdot \frac{\dz}{iz} = \frac1{2\pi i}\int_{|z|=1} \frac{(z-a)(\frac1z-\overline{a})}{(z-b)(\frac1z-\overline{b})} \cdot \frac{\dz}{z} = \)

\(\displaystyle = \frac1{2\pi i}\int_{|z|=1} \frac{(z-a)(1-\overline{a}z)}{b(1-\overline{b}z)} \left(\frac1{z-b}-\frac1z\right)\dz = \frac{(z-a)(1-\overline{a}z)}{b(1-\overline{b}z)}\Bigg|_{z=b} - \frac{(z-a)(1-\overline{a}z)}{b(1-\overline{b}z)}\Bigg|_{z=0} = \)

\(\displaystyle = \frac{(b-a)(1-\overline{a}b)}{b(1-\overline{b}b)} + \frac{a}{b} = \frac{(a-\overline{b})(\overline{a}-b)}{1-b\overline{b}}+1 = \frac{|a-b|^2}{1-|b|^2}+1. \)

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