Let \(f\) be analytic on a region \(R\) and suppose \(f'(z_0) \neq 0\) for some \(z_0 \in R\). Show that if \(C\) is a circle of sufficiently small radius centered at \(z_0\), then \[ \frac{2 \pi i}{f^{\prime}\left(z_{0}\right)}=\int_{C} \frac{d z}{f(z)-f\left(z_{0}\right)} .\]
Hint: use the inverse function theorem.
Solution
1. Local biholomorphism via the Inverse Function Theorem:
1.1. Since \(f\) is holomorphic on \(R\) and \(f'(z_0) \neq 0\), the Complex Inverse Function Theorem implies there exist open neighborhoods \(U\) of \(z_0\) and \(V\) of \(w_0 = f(z_0)\) such that \(f: U \to V\) is a biholomorphism with holomorphic inverse \(g = f^{-1}: V \to U\).
Complex Inverse Function Theorem.
1.2. The derivative of the inverse function at \(w_0\) is given by:\[ g'(w_0) = \frac{1}{f'(g(w_0))} = \frac{1}{f'(z_0)}. \]
chain rule applied to \(f(g(w)) = w\).
2. Change of variables in the contour integral:
2.1. Choose \(r > 0\) sufficiently small such that the closed disk \(\overline{D}(z_0, r) \subset U\), and let \(C = \partial D(z_0, r)\) be the circle oriented counterclockwise.
openness of \(U\).
2.2. The image \(\Gamma = f(C)\) is a simple closed curve in \(V\) winding once counterclockwise around \(w_0 = f(z_0)\), so the winding number is \(\operatorname{Ind}_\Gamma(w_0) = 1\).
conformal mapping of a small disk.
2.3. Substitute \(z = g(w)\), which gives \(dz = g'(w)\,dw\) and \(f(z) - f(z_0) = w - w_0\):\[ \int_C \frac{dz}{f(z) - f(z_0)} = \int_\Gamma \frac{g'(w)}{w - w_0} \, dw. \]
holomorphic change of variables for line integrals.
3. Evaluation via Cauchy’s Integral Formula:
3.1. Since \(g'(w)\) is holomorphic on \(V\) and \(\Gamma\) is a simple closed curve in \(V\) enclosing \(w_0\), Cauchy’s Integral Formula yields:\[ \int_\Gamma \frac{g'(w)}{w - w_0} \, dw = 2\pi i \, g'(w_0). \]
Cauchy’s Integral Formula.
3.2. Substituting \(g'(w_0) = \frac{1}{f'(z_0)}\) from 1:\[ \int_C \frac{dz}{f(z) - f(z_0)} = 2\pi i \cdot \frac{1}{f'(z_0)} = \frac{2\pi i}{f'(z_0)}. \]
1 step 2.
4. Conclusion:\(\int_C \frac{dz}{f(z) - f(z_0)} = \frac{2\pi i}{f'(z_0)}\). Q.E.D.
2 and 3.