Cauchy's Integral Formula

6.1 Statement

Theorem 6.1 (Cauchy”s Integral Formula). If $f$ is analytic on a connected domain Containing a simple closed positively oriented contour $\gamma$And $z_0$ is inside $\gamma$Then

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

Proof. Let $\gamma_\varepsilon$ be a small circle of radius $\varepsilon$ around $z_0$. Since $\frac{f(z)}{z - z_0}$ is analytic on the region between $\gamma$ and $\gamma_\varepsilon$

$\int_\gamma \frac{f(z)}{z - z_0}\, dz = \int_{\gamma_\varepsilon} \frac{f(z)}{z - z_0}\, dz$

On $\gamma_\varepsilon$: $f(z) = f(z_0) + (z - z_0)f'(\zeta)$ for some $\zeta$ between $z$ and $z_0$.

$= \int_{\gamma_\varepsilon} \frac{f(z_0)}{z - z_0}\, dz + \int_{\gamma_\varepsilon} f'(\zeta)\, dz = f(z_0) \cdot 2\pi i + 0$

Since $\int_{\gamma_\varepsilon} \frac{dz}{z - z_0} = 2\pi i$ (parameterize $z = z_0 + \varepsilon e^{i\theta}$) and $\int_{\gamma_\varepsilon} f'(\zeta)\, dz \to 0$ as $\varepsilon \to 0$ by the ML inequality. $\blacksquare$

6.2 Derivatives

Theorem 6.2 (Cauchy’s Integral Formula for Derivatives). Under the same conditions,

$f^{(n)}(z_0) = \frac{n!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz$

Proof. We proceed by induction. The base case $n = 0$ is Theorem 6.1. For the inductive step, Assume the formula holds for $n$. Using the difference quotient:

$f^{(n+1)}(z_0) = \lim_{h \to 0} \frac{f^{(n)}(z_0 + h) - f^{(n)}(z_0)}{h} = \lim_{h \to 0} \frac{n!}{2\pi i}\int_\gamma \frac{1}{h}\left[\frac{f(z)}{(z - z_0 - h)^{n+1}} - \frac{f(z)}{(z - z_0)^{n+1}}\right] dz$

$= \frac{(n+1)!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+2}}\, dz$

Where we justified passing the limit inside the integral by uniform convergence of the integrand On compact subsets. $\blacksquare$

6.3 Consequences of Cauchy’s Integral Formula

Corollary 6.3. If $f$ is analytic, then $f$ is infinitely differentiable.

This is remarkable: a single complex derivative implies the existence of all derivatives.

Corollary 6.4 (Cauchy’s Estimates). If $f$ is analytic on and inside a circle $|z - z_0| = R$ And $|f(z)| \leq M$ on the circle, then

$|f^{(n)}(z_0)| \leq \frac{n!M}{R^n}$

Proof. From the integral formula: $|f^{(n)}(z_0)| = \frac{n!}{2\pi}\left|\int_{|z-z_0|=R} \frac{f(z)}{(z-z_0)^{n+1}}\, dz\right| \leq \frac{n!}{2\pi} \cdot \frac{M}{R^{n+1}} \cdot 2\pi R = \frac{n!M}{R^n}$. $\blacksquare$

6.4 Liouville’s Theorem

Theorem 6.5 (Liouville’s Theorem). Every bounded entire function is constant.

Proof. If $|f(z)| \leq M$ for all $z$Then by Cauchy’s estimates with $R$ arbitrarily large: $|f'(z_0)| \leq \frac{M}{R} \to 0$ as $R \to \infty$. So $f'(z) = 0$ for all $z$Meaning $f$ is Constant. $\blacksquare$

Corollary 6.6. If $f$ is entire and $|f(z)| \geq M$ for all $z$ (bounded away from zero), then $f$ is constant.

Proof. $1/f$ is entire and bounded by $1/M$So constant by Liouville. $\blacksquare$

6.5 Fundamental Theorem of Algebra

Theorem 6.7 (Fundamental Theorem of Algebra). Every non-constant polynomial $p(z) \in \mathbb{C}[z]$ has a root in $\mathbb{C}$.

Proof. Suppose $p(z)$ has no root. Then $f(z) = 1/p(z)$ is entire. Since $|p(z)| \to \infty$ as $|z| \to \infty$, $f(z) \to 0$So $f$ is bounded. By Liouville’s theorem, $f$ is constant, so $p$ Is constant, a contradiction. $\blacksquare$

Corollary 6.8. Every polynomial of degree $n \geq 1$ has exactly $n$ roots in $\mathbb{C}$ Counting multiplicities.

6.6 Worked Examples: Cauchy’s Integral Formula