Cauchy’s Integral Formula

Our interest in the Cauchy-Goursat theorem isn’t the theorem’s sake, but its use in proving what might be arguably the most important theorem in introductory complex analysis—Cauchy’s integral formula.

Fix $f : \mathbb C \to \mathbb C$ .

Theorem 1 (Cauchy’s Integral Formula). Let $D \subseteq \mathbb C$ be a simply connected domain. Suppose $f$ is holomorphic on $D$ and $\partial D$ , directed anticlockwise. For any $z_0 \in D$ ,

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

Proof. Fix $z_0 \in D$ . Since $f$ is continuous at $z_0$ , for any $\epsilon > 0$ , there exists $\delta_{\epsilon} > 0$ such that

$|z-z_0| < \delta_{\epsilon} \quad \Rightarrow \quad |f(z) - f(z_0)| < \epsilon.$

Now for $R_\epsilon < \delta_\epsilon$ , $z_0 \in B(z_0, R_\epsilon) \subseteq D$ . Defining $\Gamma_\epsilon := \partial B(z_0, R_\epsilon) = r([0, 2\pi])$ , we can parameterise $\Gamma$ using $r(t) =z_0 + R_\epsilon e^{it}$ , so that $r'(t) = iR_\epsilon e^{it}$ . Then

$\begin{aligned} \oint_{\Gamma_\epsilon } \frac{f(z)}{z - z_0}\, \mathrm dz &= \int_0^{2\pi} \frac{f(z_0 + R_\epsilon e^{it})}{(z_0 + R_\epsilon e^{it}) - z_0} \cdot iR_\epsilon e^{it}\, \mathrm dt = i \cdot \int_0^{2\pi} f(z_0 + R_\epsilon e^{it})\, \mathrm dt. \end{aligned}$

Hence,

$\begin{aligned} \left| f(z_0) - \frac 1{2\pi i}\oint_{\Gamma_\epsilon } \frac{f(z)}{z - z_0}\, \mathrm dz \right| &= \left| f(z_0) - \frac 1{2\pi i} \cdot i \cdot \int_0^{2\pi} f(z_0 + Re^{it})\, \mathrm dt \right| \\ &\leq \frac 1{2\pi} \cdot \int_0^{2\pi} | f(z_0 + Re^{it}) - f(z_0) |\, \mathrm dt \\ & < \frac 1{2\pi} \int_0^{2\pi} \epsilon\, \mathrm dt = \frac 1{2\pi} \cdot \epsilon \cdot 2\pi = \epsilon. \end{aligned}$

Define $D^+ := D \cap \{z \in \mathbb C : \mathrm{Im}(z) > \mathrm{Im}(z_0)\} \backslash B(z_0, R)$ and $D^-$ similarly. Since these domains are bounded and the map $f(z)/(z- z_0)$ is holomorphic on them, by the Cauchy-Goursat theorem,

$\displaystyle \oint_{\partial D^+} \frac{f(z)}{z - z_0}\, \mathrm dz = \oint_{\partial D^-} \frac{f(z)}{z - z_0}\, \mathrm dz = 0.$

Yet, due to the cancelation features of contour integration,

$\begin{aligned} \oint_{\partial D} \frac{f(z)}{z - z_0}\, \mathrm dz &= \oint_{\partial D^+} \frac{f(z)}{z - z_0}\, \mathrm dz + \oint_{\partial D^-} \frac{f(z)}{z - z_0}\, \mathrm dz + \oint_{\Gamma_\epsilon } \frac{f(z)}{z - z_0}\, \mathrm dz = \oint_{\Gamma_\epsilon } \frac{f(z)}{z - z_0}\, \mathrm dz. \end{aligned}$

Therefore,

$\begin{aligned} \left| f(z_0) - \frac 1{2\pi i}\oint_{\partial D} \frac{f(z)}{z - z_0}\, \mathrm dz \right| &= \left| f(z_0) - \frac 1{2\pi i}\oint_{\Gamma_\epsilon } \frac{f(z)}{z - z_0}\, \mathrm dz \right| < \epsilon. \end{aligned}$

Taking $\epsilon \to 0^+$ ,

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

The implications of this theorem are exceedingly massive.

Theorem 2. If $f$ is holomorphic on $\bar D$ , then for any $z_0 \in D$ and $n \in \mathbb N$ , $f^{(n)}(z_0)$ exists and

$\displaystyle f^{(n)}(z_0) = \frac{n!}{2 \pi i} \oint_{\partial D} \frac{f(z)}{(z-z_0)^{n+1}}\, \mathrm dz,$

where $\partial D$ is directed anti-clockwise. Consequently, $f^{(n)}$ is holomorphic on $D$ for any $n$ . Heuristically,

$\displaystyle f^{(n)}(z_0) = \frac 1{2\pi i} \oint_{\partial D} \frac{\mathrm d^n}{\mathrm dz_0^n} \left( \frac{f(z)}{z-z_0} \right)\, \mathrm dz.$

Proof. Since $f$ is holomorphic at $z_0 \in D$ , find an open ball $D$ with centre $z_0$ such that $f$ is holomorphic on $\bar D$ . Therefore, for any $w \in D$ , Cauchy’s integral formula gives

$\displaystyle f(w) = \frac 1{2\pi i} \oint_{\partial D} \frac{f(z)}{z - w}\, \mathrm dz.$

In particular,

$\begin{aligned} \frac{f(z_0 + w) - f(z_0)}{w} &= \frac 1w \cdot \frac 1{2\pi i} \oint_{\partial D} f(z) \cdot \left[ \frac 1{z - (z_0 + w)} - \frac 1{z - z_0} \right]\, \mathrm dz \\ &= \frac 1{2\pi i} \oint_{\partial D} \frac {f(z)}{(z - z_0 - w)(z - z_0)}\, \mathrm dz. \end{aligned}$

Therefore,

$\begin{aligned} \frac{f(z_0 + w) - f(z_0)}{w} &- \frac 1{2\pi i} \oint_{\partial D} \frac{f(z)}{(z-z_0)^2}\, \mathrm dz \\ &= \frac{w}{2\pi i} \cdot \oint_{\partial D} \frac {f(z)}{(z - z_0 - w)(z - z_0)^2}\, \mathrm dz =: (\dagger) \cdot w. \end{aligned}$

Using the reverse triangle inequality and the ML-inequality, we can show that $(\dagger)$ is bounded, so that the right-hand side $\to 0$ as $w \to 0$ . Therefore,

$\displaystyle f'(z_0) = \lim_{w \to 0} \frac{f(z_0 + w) - f(z_0)}{w} = \frac 1{2\pi i} \oint_{\partial D} \frac{f(z)}{(z-z_0)^2}\, \mathrm dz.$

For the general case, we repeat the argument by induction, beginning with the equation

$\displaystyle \begin{aligned} \frac{f^{(n)}(z_0 + w) - f^{(n)}(z_0)}{w} &= \frac 1w \cdot \frac {n!}{2\pi i} \oint_{\partial D} f(z) \left[ \frac {1}{(z - z_0 - w)^{n+1}} - \frac {1}{(z - z_0)^{n+1}} \right]\, \mathrm dz \\ &= \frac 1{2\pi i} \oint_{\partial D} f(z) \left[ \frac {(n+1)! + O(w)}{(z - z_0 - w)^{n+1}(z - z_0)^{n+1}} \right]\, \mathrm dz, \end{aligned}$

where there exists a universal constant $C$ such that $|O(w)| \leq C \cdot |w|$ , and simplifying relevant expressions using the binomial theorem. The result will follow from taking $w \to 0$ so that $O(w) \to 0$ .

Therefore, if a function is complex-differentiable at a neighborhood of $z_0$ , it is infinitely complex-differentiable on that same neighborhood! This surprising result is unsurprisingly false in the real-differentiable setting, with the simple example of $f(x) = x|x|$ being continuously real-differentiable on $(-\epsilon, \epsilon)$ for any $\epsilon > 0$ but not twice real-differentiable there.

Likewise, it is usually the case that a function can be continuous without being differentiable; take the usual modulus function $| \cdot |$ for example. In the complex world, we have a situation where continuity does imply differentiability.

Theorem 3 (Morera’s Theorem). If $f$ is continuous on a domain $D$ and $\oint_{\Gamma} f(z)\, \mathrm dz = 0$ for every closed contour $\Gamma \subseteq D$ , then $f$ is holomorphic on $D$ .

Proof. By hypothesis, $f$ has an antiderivative $F$ on $D$ . By Theorem 2, $f = F'$ is holomorphic on $D$ .

Furthermore, we can easily bound the $n$ -th derivative of a function holomorphic at the point.

Theorem 4 (Cauchy’s Inequality). Suppose $f$ is holomorphic on $\bar B(z_0, R)$ , whose boundary $\Gamma_R$ is parametrised by the map $r : [0, 2\pi] \to \mathbb C$ , $r(t) = z_0 + Re^{it}$ . By continuity and the extreme value theorem, $M_R := \max_{z \in \Gamma_R} |f(z)| < \infty$ is well-defined. Then

$\displaystyle |f^{(n)}(z_0)| \leq \frac{n! M_R}{R^n}.$

Proof. By Cauchy’s integral formula,

$\displaystyle f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_{\Gamma_R} \frac{f(z)}{(z - z_0)^{n+1}}\, \mathrm dz.$

Since $|z - z_0| = R$ , the integrand is bounded above by

$\displaystyle \left| \frac{f(z)}{(z - z_0)^{n+1}} \right| \leq \frac{M_R}{R^{n+1}}.$

By the ML-inequality,

$\displaystyle |f^{(n)}(z_0)| \leq \frac{n!}{2\pi} \cdot \frac{M_R}{R^{n+1}} \cdot 2\pi R= \frac{n! M_R}{R^n}.$

In the real-variable case, there are many bounded functions that are differentiable on all of $\mathbb R$ . Consider the classic example $f(x) = 1/(x^2+1)$ . This works in multiple dimensions too; take the analogous function $f(\mathbf x) = 1/(\| \mathbf x \|^2 + 1)$ . But in the single complex-variable case, this scenario turns out to be an impossibility!

Theorem 5 (Liouville’s Theorem). If $f : \mathbb C \to \mathbb C$ is entire and bounded, then it must be constant i.e. there exists $w_0 \in \mathbb C$ such that $f = w_0$ . The converse is trivially true.

Proof. Fix $z_0 \in \mathbb C$ . Since $f$ is bounded, there exists $M > 0$ such that $|f| \leq M$ . By Cauchy’s inequality, given any circle with centre $z_0$ and radius $R$ ,

$\displaystyle |f'(z_0)| \leq \frac{M}{R}.$

Taking $R \to 0$ , we must have $f'(z_0) = 0$ . Since $z_0$ is arbitrary, $f' = 0$ . Since $f$ is entire, we must have $f$ being constant.

But isn’t the usual sine function bounded by $[-1, 1]$ ? In the real-world that certainly is the case; but not so with the complex case. Recall that

$\displaystyle \sin(z) := \frac{e^{iz} - e^{-iz}}{2i}.$

The use of complex numbers helps us see clearly the unboundedness: set $z = -Ri$ so that for sufficiently large $R$ ,

$\displaystyle |{\sin(z)}| = \frac{|e^R - e^{-R}|}{2} > \frac{e^R}{4}.$

Taking $R \to \infty$ demonstrates the unboundedness of $\sin : \mathbb C \to \mathbb C$ .

Corollary 1. If $f : \mathbb C \to \mathbb C$ is entire and there exists $m > 0$ such that $|f| \geq m$ , then it must be constant. The converse is trivially true.

Proof. Firstly, we note that $f \neq 0$ . Therefore, $1/f$ is entire and nonzero, and bounded above by $1/m$ . By Liouville’s theorem, $1/f$ is constant and nonzero, so that $f$ is constant as well.

Finally, we can prove the fundamental theorem of algebra (again!), that states any nonzero degree $n$ polynomial with complex coefficients will have at least one root in $\mathbb C$ .

Theorem 6 (Fundamental Theorem of Algebra). For any polynomial $p \in \mathbb C[z]$ defined by

$\displaystyle p(z) = \sum_{k=0}^n a_k z^k,\quad a_n \neq 0, \quad 0^0 := 1,$

there exist $z_1, \dots, z_n \in \mathbb C$ such that

$\displaystyle p(z) = a_n \cdot \prod_{k=1}^n (z - z_k).$

Proof. Assuming $a_n = 1$ without loss of generality. We first prove the existence of a complex root. If not, then $p(z) \neq 0$ for any $z \in \mathbb C$ . Hence, $1/p$ is entire. To prove that it is bounded, observe that

$\displaystyle \frac 1{p(z)} = \frac 1{\sum_{k=0}^n a_k z^k} = \frac 1{1+ \sum_{k=0}^n a_k / z^{n-k}} \cdot \left( \frac 1z \right)^n =: \frac{w}{z^n},$

where $w$ is bounded. Hence, taking $|z| \to \infty$ , $1/p(z) \to 0$ . This means there exists $N > 0$ such that for $|z| > N$ , $1/|p(z)| \leq 1$ . Since $|\cdot | \circ 1/p : \bar B(0, N) \to \mathbb R$ is continuous, it attains a maximum value $M$ there by the extreme value theorem. Hence, $|1/p| \leq \max \{1, M\}$ , which means that $1/p$ is bounded. By Liouville’s theorem, $1/p$ is constant, so that $p$ is constant, a contradiction.

Next time, we apply Cauchy’s integral formula to derive a just-as-powerful theorem known as the Cauchy residue theorem. We even harness its power to establish many more deep theorems in complex analysis.

—Joel Kindiak, 19 Aug 25, 2025H

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Cauchy’s Integral Formula

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