The Cauchy integral formula via homotopy

Let \(C\) be positively oriented and let \(z\) lie in its interior. Since \[ \frac{f(\zeta)}{\zeta-z} =\frac{f(\zeta)-f(z)}{\zeta-z}+\frac{f(z)}{\zeta-z}, \] set \[ h(\zeta)=\begin{cases} \dfrac{f(\zeta)-f(z)}{\zeta-z},&\zeta\ne z,\\[4pt] f'(z),&\zeta=z. \end{cases} \] Then \(h\) is holomorphic near the region swept out by a homotopy shrinking \(C\) to a sufficiently small positively oriented circle \(C_r\) centered at \(z\). By homotopy invariance of contour integrals of holomorphic functions, \[ \int_C h(\zeta)\,d\zeta=\int_{C_r}h(\zeta)\,d\zeta. \] Letting \(r\downarrow0\), the right side tends to \(0\), because \(h\) is bounded near \(z\) and the length of \(C_r\) is \(2\pi r\). Hence \[ \int_C\frac{f(\zeta)-f(z)}{\zeta-z}\,d\zeta=0. \] Also the winding number of \(C\) about \(z\) is \(1\), so \[ \int_C\frac{d\zeta}{\zeta-z}=2\pi i. \] Therefore \[ \int_C\frac{f(\zeta)}{\zeta-z}\,d\zeta=2\pi i f(z), \] which is the Cauchy integral formula.