Let \(f\) be analytic in a domain \(D\) and \(\gamma\) be a closed, piecewise smooth curve in \(D\). For any \(z_0 \in D\) not lying on \(\gamma\), show that: \[\oint_\gamma \frac{f'(z)}{z - z_0} \, dz = \oint_\gamma \frac{f(z)}{(z - z_0)^2} \, dz.\] Give a generalization of this result for higher-order derivatives and powers.
Solution
Since \(z_0\notin\gamma\), the function \[ F(z)=\frac{f(z)}{z-z_0} \] is holomorphic on a neighborhood of the curve. Hence \[ F'(z)=\frac{f'(z)}{z-z_0}-\frac{f(z)}{(z-z_0)^2}, \] and the integral of \(F'\) around the closed curve is zero. Therefore \[ \oint_\gamma\frac{f'(z)}{z-z_0}\,dz = \oint_\gamma\frac{f(z)}{(z-z_0)^2}\,dz. \]
More generally, for integers \(k,m\ge1\), \[ \left(\frac{f^{(k-1)}(z)}{(z-z_0)^m}\right)' = \frac{f^{(k)}(z)}{(z-z_0)^m} -m\frac{f^{(k-1)}(z)}{(z-z_0)^{m+1}}. \] Thus \[ \oint_\gamma\frac{f^{(k)}(z)}{(z-z_0)^m}\,dz =m\oint_\gamma\frac{f^{(k-1)}(z)}{(z-z_0)^{m+1}}\,dz. \] Iterating gives \[ \boxed{ \oint_\gamma\frac{f^{(k)}(z)}{(z-z_0)^m}\,dz = \frac{(m+k-1)!}{(m-1)!} \oint_\gamma\frac{f(z)}{(z-z_0)^{m+k}}\,dz }. \]