Proof of the Cauchy Integral Formula

We present a complete proof of the Cauchy integral formula, which states that a holomorphic function inside a simple closed contour is completely determined by its boundary values.

The Statement

Theorem3.12Cauchy integral formula (restated)

Let $D$ be a simply connected domain, $f: D \to \mathbb{C}$ holomorphic, and $\gamma$ a positively oriented simple closed contour in $D$ . For every $z_0$ in the interior of $\gamma$ :

$f(z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - z_0}\,dz.$

Complete Proof

Proof

Step 1: Deformation to a small circle.

Let $\varepsilon > 0$ be small enough that $\overline{D}(z_0, \varepsilon)$ is contained in the interior of $\gamma$ . The function $g(z) = f(z)/(z - z_0)$ is holomorphic on the region between $\gamma$ and $C_\varepsilon = \{|z - z_0| = \varepsilon\}$ . By Cauchy's theorem for multiply connected domains:

$\oint_\gamma \frac{f(z)}{z - z_0}\,dz = \oint_{C_\varepsilon} \frac{f(z)}{z - z_0}\,dz.$

Step 2: Evaluation on the small circle.

On $C_\varepsilon$ , write $z = z_0 + \varepsilon e^{i\theta}$ , so $dz = i\varepsilon e^{i\theta}\,d\theta$ and $z - z_0 = \varepsilon e^{i\theta}$ :

$\oint_{C_\varepsilon} \frac{f(z)}{z - z_0}\,dz = \int_0^{2\pi} \frac{f(z_0 + \varepsilon e^{i\theta})}{\varepsilon e^{i\theta}} \cdot i\varepsilon e^{i\theta}\,d\theta = i\int_0^{2\pi} f(z_0 + \varepsilon e^{i\theta})\,d\theta.$

Step 3: Decomposition.

We write

$\oint_{C_\varepsilon} \frac{f(z)}{z-z_0}\,dz = f(z_0) \oint_{C_\varepsilon} \frac{dz}{z-z_0} + \oint_{C_\varepsilon} \frac{f(z) - f(z_0)}{z - z_0}\,dz.$

The first integral equals $2\pi i$ (direct computation).

Step 4: The remainder vanishes.

For the second integral, since $f$ is continuous at $z_0$ , for any $\delta > 0$ there exists $\varepsilon_0$ such that $|f(z) - f(z_0)| < \delta$ when $|z - z_0| < \varepsilon_0$ . By the ML inequality, for $\varepsilon < \varepsilon_0$ :

$\left|\oint_{C_\varepsilon} \frac{f(z) - f(z_0)}{z - z_0}\,dz\right| \leq \frac{\delta}{\varepsilon} \cdot 2\pi\varepsilon = 2\pi\delta.$

Since $\delta$ is arbitrary, this integral is zero.

Step 5: Conclusion.

Combining Steps 1--4:

$\oint_\gamma \frac{f(z)}{z-z_0}\,dz = 2\pi i \cdot f(z_0) + 0 = 2\pi i \cdot f(z_0).$

Dividing by $2\pi i$ yields the Cauchy integral formula. $\blacksquare$

■

Key Consequences

RemarkImmediate corollaries

From this proof, we can extract several important consequences:

Analyticity: Differentiating under the integral sign (justified by uniform convergence) gives $f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-z_0)^{n+1}}\,dz$ , proving that holomorphic functions are infinitely differentiable.
Cauchy's inequality: Setting $\gamma = \{|z-z_0| = R\}$ and using the ML inequality on the derivative formula yields $|f^{(n)}(z_0)| \leq n! M / R^n$ .
Liouville's theorem: If $f$ is entire and bounded by $M$ , Cauchy's inequality gives $|f'(z_0)| \leq M/R \to 0$ as $R \to \infty$ , so $f' \equiv 0$ and $f$ is constant.
Power series representation: The geometric series expansion $\frac{1}{z-z_0} = \sum_{n=0}^\infty \frac{(w-z_0)^n}{(z-z_0)^{n+1}}$ substituted into the Cauchy formula yields the Taylor series $f(w) = \sum_{n=0}^\infty a_n(w-z_0)^n$ .

ExampleFundamental Theorem of Algebra from Liouville

If $p(z) = a_n z^n + \cdots + a_0$ with $a_n \neq 0$ and $n \geq 1$ has no zeros, then $1/p(z)$ is entire. Since $|p(z)| \to \infty$ as $|z| \to \infty$ , $1/p$ is bounded, so by Liouville's theorem $1/p$ is constant — a contradiction. Therefore $p$ must have at least one zero.