State and prove the Cauchy integral formula for holomorphic functions.
Solution
Statement of the Cauchy Integral Formula
Theorem: Let \(\Omega \subseteq \mathbb{C}\) be an open domain, and let \(f: \Omega \to \mathbb{C}\) be a holomorphic function. Let \(\gamma\) be a positively oriented, simple closed rectifiable curve in \(\Omega\) whose interior \(\text{Int}(\gamma)\) is contained in \(\Omega\). Then for every point \(z_0 \in \text{Int}(\gamma)\): \[f(z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - z_0} \, dz.\]
Proof
1. Isolate the singularity at \(z_0\) using a small circle \(C_\varepsilon\).
1.1. Since \(z_0 \in \text{Int}(\gamma)\) and \(\text{Int}(\gamma)\) is open, there exists \(\varepsilon_0 > 0\) such that the closed disk \(\overline{D}(z_0, \varepsilon_0) \subset \text{Int}(\gamma)\).Proof: \(\text{Int}(\gamma)\) is an open neighborhood of \(z_0\).
1.2. For any \(\varepsilon \in (0, \varepsilon_0)\), let \(C_\varepsilon\) denote the circle \(\{z \in \mathbb{C} : |z - z_0| = \varepsilon\}\), oriented counterclockwise.Proof: This is a definition.
1.3. The function \(g(z) \coloneqq \frac{f(z)}{z - z_0}\) is holomorphic in the region \(\Omega' = \text{Int}(\gamma) \setminus \overline{D}(z_0, \varepsilon)\).Proof: Quotient of holomorphic functions with non-vanishing denominator on \(\Omega'\).
1.4. By Cauchy’s Integral Theorem for multiply connected domains: \[\oint_\gamma \frac{f(z)}{z - z_0} \, dz = \oint_{C_\varepsilon} \frac{f(z)}{z - z_0} \, dz.\] Proof: The boundary of \(\Omega'\) is \(\gamma - C_\varepsilon\), and \(\oint_{\partial \Omega'} g(z)\,dz = 0\).
1.5. Q.E.D.
2. Evaluate the constant-numerator integral on \(C_\varepsilon\).
2.1. Parametrize \(C_\varepsilon\) by \(z(\theta) = z_0 + \varepsilon e^{i\theta}\) for \(\theta \in [0, 2\pi]\), so \(dz = i \varepsilon e^{i\theta} d\theta\).Proof: Definition of circular parametrization.
2.2. The integral of \(\frac{1}{z - z_0}\) evaluates to: \[\oint_{C_\varepsilon} \frac{dz}{z - z_0} = \int_0^{2\pi} \frac{i \varepsilon e^{i\theta}}{\varepsilon e^{i\theta}} \, d\theta = i \int_0^{2\pi} d\theta = 2\pi i.\] Proof: Direct cancellation of \(\varepsilon e^{i\theta}\).
2.3. Multiplying by the constant \(f(z_0)\): \[\oint_{C_\varepsilon} \frac{f(z_0)}{z - z_0} \, dz = f(z_0) \oint_{C_\varepsilon} \frac{dz}{z - z_0} = 2\pi i f(z_0).\] Proof: Linearity of integration.
2.4. Q.E.D.
3. Split the integral into the value at \(z_0\) and an error term.
3.1. Subtracting the identity from <1>2.<2>3: \[\oint_\gamma \frac{f(z)}{z - z_0} \, dz - 2\pi i f(z_0) = \oint_{C_\varepsilon} \frac{f(z) - f(z_0)}{z - z_0} \, dz.\] Proof: Follows from <1>1.<2>4 and <1>2.<2>3.
3.2. Q.E.D.
4. Show that the error term vanishes as \(\varepsilon \to 0^+\).
4.1. Since \(f\) is holomorphic at \(z_0\), \(f\) is continuous at \(z_0\): for any \(\delta > 0\), there exists \(\varepsilon > 0\) such that \(|z - z_0| = \varepsilon \implies |f(z) - f(z_0)| \leq \delta\).Proof: Continuity of \(f\) at \(z_0\).
4.2. On \(C_\varepsilon\), the integrand is bounded by: \[\left| \frac{f(z) - f(z_0)}{z - z_0} \right| = \frac{|f(z) - f(z_0)|}{\varepsilon} \leq \frac{\delta}{\varepsilon}.\] Proof: \(|z - z_0| = \varepsilon\).
4.3. The length of \(C_\varepsilon\) is \(2\pi \varepsilon\).Proof: Circumference of circle of radius \(\varepsilon\).
4.4. By the \(ML\)-inequality: \[\left| \oint_{C_\varepsilon} \frac{f(z) - f(z_0)}{z - z_0} \, dz \right| \leq \frac{\delta}{\varepsilon} \cdot (2\pi \varepsilon) = 2\pi \delta.\] Proof: \(ML\)-inequality for contour integrals.
4.5. Since the LHS of <1>3.<2>1 is independent of \(\varepsilon\), and \(\delta > 0\) can be made arbitrarily small by choosing \(\varepsilon\) small, the error integral must be identically 0: \[\oint_\gamma \frac{f(z)}{z - z_0} \, dz - 2\pi i f(z_0) = 0.\] Proof: A non-negative quantity bounded by \(2\pi\delta\) for all \(\delta > 0\) is 0.
4.6. Q.E.D.
5. Conclusion.
5.1. Dividing by \(2\pi i\) gives: \[f(z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - z_0} \, dz.\] Proof: Rearrangement of <1>4.<2>5.
5.2. Q.E.D.