KindiakMath

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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