Let \(f\) be a holomorphic function on the disk \(D_{R_0}(0) \subset \mathbb{C}\) of radius \(R_0 > 0\).
Prove that whenever \(0 < R < R_0\) and \(|z| < R\), \[ f(z) = \frac{1}{2\pi} \int_0^{2\pi} f(R e^{i\varphi}) \operatorname{Re}\left( \frac{R e^{i\varphi} + z}{R e^{i\varphi} - z} \right) d\varphi. \]
Show that for \(z = r \in [0, R)\) (and in general with \(\gamma = \varphi - \theta\) where \(z = r e^{i\theta}\)): \[ \operatorname{Re}\left( \frac{R e^{i\gamma} + r}{R e^{i\gamma} - r} \right) = \frac{R^2 - r^2}{R^2 - 2Rr\cos\gamma + r^2}. \]
Solution
For part (b), let \(z=re^{i\theta}\) and put \(\gamma=\varphi-\theta\). Rotating by \(e^{-i\theta}\) reduces the computation to real \(r\). Then \[ \frac{Re^{i\gamma}+r}{Re^{i\gamma}-r} = \frac{(Re^{i\gamma}+r)(Re^{-i\gamma}-r)}{|Re^{i\gamma}-r|^2}. \] The numerator has real part \(R^2-r^2\), while \[ |Re^{i\gamma}-r|^2=R^2-2Rr\cos\gamma+r^2. \] Hence \[ \operatorname{Re}\frac{Re^{i\gamma}+r}{Re^{i\gamma}-r} = \frac{R^2-r^2}{R^2-2Rr\cos\gamma+r^2}. \]
For part (a), first suppose \(0<|z|<R\) and set \[ w=\frac{R^2}{\overline z}. \] Then \(|w|>R\). By Cauchy’s formula and Cauchy’s theorem, \[ f(z)=\frac1{2\pi i}\int_{|\zeta|=R}f(\zeta) \left(\frac1{\zeta-z}-\frac1{\zeta-w}\right)d\zeta. \] On \(|\zeta|=R\), \[ \zeta\left(\frac1{\zeta-z}-\frac1{\zeta-w}\right) = \frac{\zeta}{\zeta-z}+\frac{\overline z}{\overline\zeta-\overline z} = \operatorname{Re}\frac{\zeta+z}{\zeta-z}. \] With \(\zeta=Re^{i\varphi}\) and \(d\zeta=i\zeta\,d\varphi\), this gives \[ f(z)=\frac1{2\pi}\int_0^{2\pi} f(Re^{i\varphi}) \operatorname{Re}\left(\frac{Re^{i\varphi}+z}{Re^{i\varphi}-z}\right)d\varphi. \] If \(z=0\), the formula is exactly the usual mean-value formula, so it holds for every \(|z|<R\).