Cauchy's Integral Theorem and Formula

Proof of Theorem 1 (Goursat's approach): We first prove this for a triangle $\Delta$, then extend to general curves. Step 1: The case of a triangle. Let $T$ be a closed triangular path in $U$. We must show $\int_T f(z) \, dz = 0$. Subdivide $T$ into four congruent triangular subpaths $T^{(1)}_1, T^{(1)}_2, T^{(1)}_3, T^{(1)}_4$ by connecting midpoints. By the orientation convention, the integrals over interior edges cancel, so

At least one subtriangle satisfies $\left|\int_{T^{(1)}_j} f(z) \, dz\right| \geq \frac{1}{4}\left|\int_T f(z) \, dz\right|$. Call this $T^{(1)}$. Iterate this process: construct a sequence of triangular paths $T \supseteq T^{(1)} \supseteq T^{(2)} \supseteq \cdots$ where each $T^{(n+1)}$ is a sub-triangle of $T^{(n)}$ with side length $2^{-n}$ times that of $T$, and

By the nested interval theorem, there exists a unique point $a \in \bigcap_{n=1}^\infty T^{(n)}$. Since $f$ is holomorphic at $a$, we have $f(z) = f(a) + f'(a)(z-a) + \psi(z)(z-a)$ where $\lim_{z \to a} \psi(z) = 0$. For the constant and linear terms, $\int_{T^{(n)}} [f(a) + f'(a)(z-a)] \, dz = 0$ (integration of polynomials around closed curves). Thus $\int_{T^{(n)}} f(z) \, dz = \int_{T^{(n)}} \psi(z)(z-a) \, dz$. Let $L_n$ be the perimeter of $T^{(n)}$. Then $L_n = 2^{-n} L$ where $L$ is the perimeter of $T$. For $\epsilon > 0$, choose $\delta > 0$ such that $|z - a| < \delta \implies |\psi(z)| < \epsilon$. For $n$ large enough that the diameter of $T^{(n)}$ is less than $\delta$, we have for all $z$ on $T^{(n)}$: $|\psi(z)| < \epsilon$ and $|z - a| \leq L_n$. Thus:

But also $\left|\int_{T^{(n)}} f(z) \, dz\right| \geq 4^{-n} \left|\int_T f(z) \, dz\right|$. Therefore $\left|\int_T f(z) \, dz\right| \leq \epsilon L^2$ for all $\epsilon > 0$, implying $\int_T f(z) \, dz = 0$. Step 2: General closed curves in simply connected domains. Since $U$ is simply connected, any closed curve $\gamma$ in $U$ bounds a region contained in $U$. By Green's theorem considerations or triangulation arguments, we can approximate the integral over $\gamma$ by sums of integrals over triangular paths, each of which vanishes by Step 1. Taking limits establishes the result for general piecewise-$C^1$ curves. (A more rigorous approach uses that in a simply connected domain, $f$ has a primitive function $F$ with $F' = f$, constructed via path integration. Then $\oint_\gamma f(z) \, dz = F(\gamma(b)) - F(\gamma(a)) = 0$ since $\gamma$ is closed.) □

Proof of Theorem 2 (Cauchy's Integral Formula): We may assume $n(\gamma, a) = 1$ by linearity (the general case follows by considering connected components). Step 1: Consider the function $g(z) = \frac{f(z) - f(a)}{z - a}$ for $z \neq a$. We can extend $g$ to be continuous at $a$ by setting $g(a) = f'(a)$. This extended function is holomorphic on $U \setminus \{a\}$. Step 2: Let $\epsilon > 0$ be small enough that the circle $C_\epsilon(a) = \{z : |z - a| = \epsilon\}$ lies in $U$ and in the interior of $\gamma$. By deformation of contours (using Cauchy's theorem in the region between $\gamma$ and $C_\epsilon(a)$),

Step 3: Parametrize $C_\epsilon(a)$ as $z = a + \epsilon e^{i\theta}$, $\theta \in [0, 2\pi]$. Then:

As $\epsilon \to 0$, by continuity of $f$, this approaches $i \int_0^{2\pi} f(a) \, d\theta = 2\pi i f(a)$. But the integral is independent of $\epsilon$ by Step 2, so

Proof of Theorem 3 (Derivatives): We prove this by induction. The case $n = 0$ is Theorem 2. Assume the formula holds for $n$. We must show:

Consider the difference quotient:

Algebraically,

Using the identity $A^{n+1} - B^{n+1} = (A-B)(A^n + A^{n-1}B + \cdots + B^n)$ with $A = z-a$ and $B = z-a-h$:

As $h \to 0$, this sum has $n+1$ terms each approaching $(z-a)^n$, so the ratio approaches $\frac{(n+1)}{(z-a)^{n+2}}$. By the dominated convergence theorem for path integrals (justified by uniform bounds on compact sets away from $a$),