Cauchy's integral formula | Complex Analysis | Mathematics

Theorem 2.1 (Cauchy’s integral formula). Let \(f\) be analytic in a simply connected domain \(D\), and let \(\Gamma \) be a simple closed positively oriented contour in \(D\). For every \(z_0\) inside \(\Gamma \),

\[ f(z_0) = \frac {1}{2\pi i} \oint _\Gamma \frac {f(z)}{z - z_0} \dd z. \]

Source: [1, ch. 4]

Proof. Let \(C_r\) be the circle \(\abs {z - z_0} = r\), with \(r\) small enough that \(C_r\) lies inside \(\Gamma \). The integrand is analytic everywhere except at \(z_0\), so by Theorem 1.3

\[ \oint _\Gamma \frac {f(z)}{z - z_0} \dd z = \oint _{C_r} \frac {f(z)}{z - z_0} \dd z = f(z_0) \oint _{C_r} \frac {\dd z}{z - z_0} + \oint _{C_r} \frac {f(z) - f(z_0)}{z - z_0} \dd z. \]

Parametrising \(z = z_0 + re^{it}\), \(0 \leq t \leq 2\pi \), the first integral is \(\int _0^{2\pi } i \dd t = 2\pi i\). Let \(E_r\) be the second. On \(C_r\) its integrand is at most \(\max _{C_r} \abs {f(z) - f(z_0)} / r\) in modulus, so by Theorem 1.1

\[ \abs {E_r} \leq 2\pi \max _{\abs {z - z_0} = r} \abs {f(z) - f(z_0)} \longrightarrow 0 \quad \text {as } r \to 0, \]

by continuity of \(f\) at \(z_0\). But \(E_r = \oint _\Gamma \frac {f(z)}{z - z_0} \dd z - 2\pi i f(z_0)\) does not depend on \(r\), so \(E_r = 0\).