Category: Theory

Laplace Inversion

In this post, we apply the culmination of our discussions in complex analysis to sufficiently legitimise the inverse Laplace transform for polytechnic courses in differential equations, taking heavy inspiration from this thread. Unfortunately, Fourier analysis only gets us the desired intuition for the Laplace inversion formula—we’ll need complex analysis to prove it properly.

For any function $f : [0, \infty) \to \mathbb C$ , recall the Laplace transform $\mathcal L\{f\} \equiv F$ defined by

$\displaystyle F(s) := \int_0^\infty f(t)e^{-st}\, \mathrm dt$

whenever the integral converges. Recall also the Fourier transform $\hat f$ of $f$ defined by

$\displaystyle \hat f(\xi) := \frac 1{2\pi} \int_{0}^{\infty} f(x)e^{-i \xi x}\,\mathrm dx.$

whenever its integral converges. The Fourier inversion theorem tells us that if $f$ is continuous and Lebesgue-integrable, then $\hat f$ exists and

$\displaystyle f(x) = \int_{-\infty}^{\infty} \hat f(\xi) e^{i \xi x}\, \mathrm d\xi.$

Theorem 1. Suppose $f$ is differentiable and there exists some $g : [0, \infty) \to \mathbb C$ such that

$\displaystyle f(u) = f(0) + \int_0^u g(v)\, \mathrm dv,$

so that $f' = g$ , and there exists $\sigma \in \mathbb R$ such that $f(t)e^{-\sigma t}$ and $g(t)e^{-\sigma t}$ are Lebesgue-integrable. Then for any $t > 0$ ,

$\displaystyle f(t) = \frac 1{2 \pi i} \cdot \lim_{R \to \infty} \oint_{\Gamma_R} F(s) e^{st}\, \mathrm ds,$

where $\Gamma_R = r( [-R, R] )$ , $r(t) = \sigma + i t$ . In particular, if $F = G$ , then $f = g$ , yielding the required Laplace inversion.

Remark 1. The intuition for this formula is setting $s = i\xi$ to recover the inverse Fourier transform. The technicality involves the Fourier’s limited effectiveness on only Lebesgue-integrable functions, and how even approximating techniques do not work since the powerful integration convergence theorems still require the baseline condition of integrability (or sadly, un-guaranteed non-negativity).

Proof. Firstly, we define the bounded function $S : \mathbb R \to \mathbb C$ by

$\displaystyle S(u) = \frac 12 + \frac 1{\pi} \int_0^u \frac{\sin (v)}{v}\, \mathrm dt \quad \Rightarrow \quad S'(u) = \frac 1{\pi} \cdot \frac{\sin(u)}{u}.$

Using the Dirichlet integral,

$\displaystyle \lim_{R \to \infty} S(Ru) = \begin{cases} 1, & u > 0, \\ 1/2, & u = 0, \\ 0, & u < 0. \end{cases}$

For each $u \neq 0$ , $S(Ru) \to H(u)$ as $R \to \infty$ , where $H := \mathbb I_{[0, \infty)}$ denotes the unit step function. To simplify the contour integral, parameterise and expand the Laplace transform to obtain

$\begin{aligned} \frac 1{2\pi i} \oint_{\Gamma_R} F(s) e^{ts}\, \mathrm ds &= \frac 1{2\pi i} \cdot \int_{-R}^R F(\sigma + i \xi) e^{t (\sigma + i \xi)}\cdot i \, \mathrm d\xi \\ &= \frac 1{2\pi} \cdot \int_{-R}^R \left( \int_0^\infty f(u) e^{-(\sigma + i \xi)u} \, \mathrm du \right) e^{t (\sigma + i \xi)} \, \mathrm d\xi \\ &= \frac 1{2\pi} \cdot \int_0^\infty f(u) \int_{-R}^R e^{-(\sigma + i \xi)u}e^{t (\sigma + i \xi)} \, \mathrm d\xi \, \mathrm du \\ &= \frac 1{2\pi} \cdot \int_0^\infty f(u) e^{-\sigma (u-t)} \int_{-R}^R e^{i \xi (t-u)} \, \mathrm d\xi \, \mathrm du \\ &= \frac 1{2\pi} \cdot e^{\sigma t} \cdot \int_0^\infty f(u) e^{-\sigma u}\cdot \frac{2 \cdot \sin (u-t)R}{(u-t)} \, \mathrm dt \\ &= \frac 1{2\pi} \cdot 2\pi e^{\sigma t} \cdot \int_0^\infty f(u) e^{-\sigma u}\cdot S'(R(u-t)) \cdot R \, \mathrm du \\ &= e^{\sigma t} \cdot \int_0^\infty f(u) e^{-\sigma u}\cdot S'(R(u-t)) \cdot R \, \mathrm du, \end{aligned}$

where we exchanged the integrals using Fubini’s theorem. By hypothesis, $f(u)e^{-\sigma u}$ and $h(u):= (f(u)e^{-\sigma u})' = (f'(u) - \sigma f(u)) e^{-\sigma u}$ are Lebesgue-integrable, so that we can integrate by parts:

$\begin{aligned} & \int_0^\infty f(u) e^{-\sigma u}\cdot S'(R(u-t)) \cdot R \, \mathrm du \\ &= \left[ f(u) e^{-\sigma u} \cdot S(R(u-t))\right]_0^\infty - \int_0^\infty h(u) \cdot S(R(u-t))\, \mathrm du \\ &= (0 - 0) - \int_0^\infty h(u) \cdot S(R(u-t))\, \mathrm du \\ &= - \int_0^\infty h(u) \cdot S(R(u-t))\, \mathrm du. \end{aligned}$

Taking $R \to \infty$ , apply the dominated convergence theorem to obtain

$\begin{aligned} \lim_{R \to \infty} \frac 1{2\pi i} \oint_{\Gamma_R} F(s) e^{ts}\, \mathrm ds &= -e^{\sigma t} \cdot \lim_{R \to \infty} \int_0^\infty h(u) \cdot S(R(u-t))\, \mathrm du \\ &= -e^{\sigma t} \cdot \int_0^\infty h(u) \cdot \lim_{R \to \infty} S(R(u-t))\, \mathrm du \\ &= -e^{\sigma t} \cdot \int_0^\infty h(u) \cdot H(u-t)\, \mathrm du \\ &= -e^{\sigma t} \cdot \int_t^\infty h(u) \, \mathrm du \\ &= -e^{\sigma t} \cdot \left[ f(u) e^{-\sigma u} \right]_t^\infty \\ &= - e^{\sigma t} \cdot (0 - f(t) e^{-\sigma t}) = f(t),\end{aligned}$

as required.

With that, we finally complete our discussion on complex analysis!

—Joel Kindiak, 29 Aug 25, 1158H

February 10, 2026
Fourier Inversion

Having defined the Fourier transform of an integrable function $f : \mathbb R \to \mathbb C$ by $\mathcal F\{f\} := \hat f : \mathbb R \to \mathbb C$ ,

$\displaystyle \hat f(\xi) := \frac 1{2\pi}\int_{-\infty}^{\infty} f(x) e^{-i \xi x}\, \mathrm dx,$

can we take its inverse? That is, is there a reasonable definition of $\mathcal F^{-1}$ such that $\mathcal F^{-1} \circ \mathcal F = \mathcal F \circ \mathcal F^{-1} = \mathrm{id}$ ?

We also need to consider a suitable collection $\mathcal S$ of functions $\mathbb R \to \mathbb C$ so that, in more technical parlance, $\mathcal F : \mathcal S \to \mathcal S$ is bijective with inverse $\mathcal F^{-1} : \mathcal S \to \mathcal S$ , but we will consider that later on.

Let $f : \mathbb R \to \mathbb C$ be integrable with Fourier transform $\hat f$ . Denote the set of such functions by $L^1(\mathbb R)$ . Furthermore, denote the set of functions $\mathbb R \to \mathbb C$ continuous on $\mathbb R$ by $C(\mathbb R)$ .

Lemma 1. For any $f \in L^1(\mathbb R)$ , $\hat f \in C(\mathbb R)$ .

Proof. For any $\xi_n \to \xi$ , the sequence of functions $\{ f(x) e^{-i \xi_n x} \}$ is dominated by the function $|f(x)|$ and converges pointwise to $f(x) e^{-i \xi x}$ . By the dominated convergence theorem,

$\begin{aligned} \lim_{n \to \infty} \hat f(\xi_n) &= \lim_{n \to \infty} \frac 1{2\pi}\int_{-\infty}^{\infty} f(x) e^{-i \xi_n x}\, \mathrm dx \\ &= \lim_{n \to \infty} \frac 1{2\pi}\int_{-\infty}^{\infty} f(x) e^{-i \xi_n x}\, \mathrm dx \\ &= \frac 1{2\pi}\int_{-\infty}^{\infty} \lim_{n \to \infty} f(x) e^{-i \xi_n x}\, \mathrm dx \\ &= \frac 1{2\pi}\int_{-\infty}^{\infty} f(x) e^{-i \xi x}\, \mathrm dx = \hat f(\xi). \end{aligned}$

Therefore, $\hat f \in C(\mathbb R)$ .

Following this article, we will prove the Fourier inversion formula.

Theorem 1 (Fourier Inversion Formula). For any $f \in L^1(\mathbb R) \cap C(\mathbb R)$ such that $\hat f \in L^1(\mathbb R)$ ,

$\displaystyle f(x) := \int_{-\infty}^{\infty} \hat f(\xi) e^{i \xi x}\, \mathrm d\xi.$

Proof. We first use the dominated convergence theorem to create an epsilon of room, and then use Fubini’s theorem to interchange the integrals:

$\begin{aligned} \int_{-\infty}^{\infty} \hat f(\xi) e^{i \xi x}\, \mathrm d\xi &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} \underbrace{ \phantom{.} e^{-\epsilon^2 \xi^2/2 + i \xi x} \phantom{.}}_{g_x(\xi)}\cdot \hat f(\xi) \, \mathrm d\xi \\ &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} g_x(\xi) \cdot \int_{-\infty}^{\infty} f(y) e^{-i \xi y}\, \mathrm dy \, \mathrm d\xi \\ &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} f(y) \int_{-\infty}^{\infty} g_x(\xi) e^{-i \xi y}\, \mathrm d\xi\, \mathrm dy \\ &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} f(y) \cdot \hat g_x(y) \, \mathrm dy .\end{aligned}$

By the Fourier transform of the Gaussian function $h(\xi) = e^{-\epsilon^2 \xi^2/2 }$ so that $\hat h(y) = \frac{1}{\epsilon \sqrt{2 \pi}} e^{-y^2/2\epsilon^2} =: \phi(y/\epsilon)/\epsilon$ , where $\phi$ denotes the probability density function of $\mathcal N(0, 1)$ ,

$\begin{aligned} \hat g_x (y) &= \int_{-\infty}^{\infty} e^{-\epsilon^2 \xi^2/2 + i \xi x} e^{-i\xi y}\, \mathrm d\xi \\ &= \int_{-\infty}^{\infty} e^{-\epsilon^2 \xi^2/2 } e^{-i\xi (y-x)}\, \mathrm d\xi \\ &= \hat h(y-x) = \frac{1}{\epsilon } \cdot \phi \left( \frac{x-y}{\epsilon} \right). \end{aligned}$

Now denote $\phi_{\epsilon}(\cdot) := \hat h = \phi(\cdot/\epsilon)/\epsilon = \hat h$ so that $\hat g_x(y) = \phi_{\epsilon}(x-y)$ , and we have

$\begin{aligned} \int_{-\infty}^{\infty} \hat f(\xi) e^{i \xi x}\, \mathrm d\xi &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} f(y) \cdot \phi_{\epsilon}(x-y) \, \mathrm dy.\end{aligned}$

All that remains is to prove that the limit approaches $f(x)$ as $\epsilon \to 0$ . Define

$\begin{aligned}\psi(\epsilon) := \int_{-\infty}^{\infty} f(y) \cdot \phi_{\epsilon}(x-y) \, \mathrm dy - f(x).\end{aligned}$

Since $\phi_{\epsilon}(x-\cdot)$ is the probability density function of some normal distribution, it has a total integral of $1$ , so that

$\begin{aligned}\psi(\epsilon) = \int_{-\infty}^{\infty} (f(y) - f(x)) \cdot \phi_{\epsilon}(x-y) \, \mathrm dy .\end{aligned}$

Fix $\eta > 0$ . Since $f$ is continuous at $x$ , for any $k > 0$ , there exists $\delta > 0$ such that

$\displaystyle |x-y| < \delta \quad \Rightarrow \quad |f(x) - f(y)| < k \cdot \eta.$

Therefore,

$\begin{aligned} | \psi(\epsilon)| &\leq \left( \int_{|x-y| < \delta} + \int_{|x-y| \geq \delta} \right) |f(y) - f(x)| \cdot |\phi_{\epsilon}(x-y)| \, \mathrm dy \\ &\leq k \cdot \eta \cdot 2\delta + \phi_\epsilon(\delta) \cdot \int_{\mathbb R} |f(y) | \, \mathrm dy + 2|f(x)| \cdot \Phi_\epsilon(-\delta), \end{aligned}$

where $\Phi_\epsilon$ denotes the cumulative distribution function of $\mathcal N(0, \epsilon)$ . For sufficiently small $\delta > 0$ , all three summands can be bounded above by $\eta/3$ , so that $|\psi(\epsilon)| < \eta$ . In particular, setting $\eta = \epsilon$ , $\phi(\epsilon) \to 0$ as $\epsilon \to 0^+$ , as required:

$\begin{aligned} \int_{-\infty}^{\infty} \hat f(\xi) e^{i \xi x}\, \mathrm d\xi &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} f(y) \cdot \phi_{\epsilon}(x-y) \, \mathrm dy = f(x).\end{aligned}$

This theorem is arguably one of the most important ones in Fourier analysis, though not the most powerful, since $\hat f$ may not be Lebesgue-integrable even if $f$ is (e.g. $f = \mathbb I_{[-1,1]}$ implies $\hat f(\xi) = \sin(\xi)/\pi \xi$ ). Using Theorem 1 to define

$\mathcal G : \mathcal F(L^1(\mathbb R) \cap C(\mathbb R)) \cap L^1(\mathbb R) \to L^1(\mathbb R) \cap C(\mathbb R)$

by $\mathcal G\{\hat f\} = f$ , we have proven that $(\mathcal G \circ \mathcal F)\{f\} = f$ for suitable $f$ .

In a limited sense then, $\mathcal G = \mathcal F^{-1}$ , although a more proper proof requires bijectivity, which we are far from a complete proof, but thankfully do not need at the moment. What’s perhaps somewhat more surprising is the connection we recover with inverse Laplace transforms.

Theorem 2. For continuous Lebesgue-integrable $f,g : [0,\infty) \to \mathbb C$ , if $\mathcal L\{f\} = \mathcal L\{g\}$ is defined on $\gamma + i \mathbb R$ , then $f = g$ . Therefore, for any $F = \mathcal L\{f\}$ defined on $i \mathbb R$ , $f = \mathcal L^{-1}\{F\}$ is well-defined.

Proof. Suppose $F:=\mathcal L\{f\} = \mathcal L\{g\} =: G$ . We first prove the $\gamma = 0$ case. Given $\xi \in \mathbb R$ such that $F(i \xi)$ is well-defined,

$\displaystyle \hat f(\xi) = \frac{1}{2\pi} \cdot \mathcal L \{ f\}(i\xi) = \frac 1{2\pi} \cdot F(i \xi).$

Similarly, $\hat g(\xi) = \frac 1{2\pi} \cdot G(i \xi)$ . By Fourier inversion, $\hat f = \hat g$ implies $f = g$ .

For the $\gamma \neq 0$ case, we note that $F(\gamma + \cdot) = G(\gamma + \cdot)$ is defined on $i \mathbb R$ . By the previous result and the shift theorems,

$\displaystyle e^{-\gamma t} \cdot f(t) = e^{-\gamma t}\cdot g(t) \quad \Rightarrow \quad f = g.$

What if $f$ is not integrable, say the simplest case $f = \mathbb I_{[0, \infty)}$ ? As much as I wanted to end complex analysis here, it would do the theory of inverse Laplace transforms injustice. We plug in this hole in our final post, next time.

—Joel Kindiak, 27 Aug 25, 1519H

February 3, 2026
The Fourier Transform

Let’s finally define the Fourier transform, carefully, as usual. In the Fourier series expansion $S_n(f)$ , we have

$\displaystyle (S_n(f))(t) = \sum_{k=-\infty}^{\infty} c_k e^{ikt},\quad c_k := \frac 1{2\pi}\int_{-\pi}^{\pi} f(t) e^{-ikt}\, \mathrm dt.$

If we regard $k$ as a variable, then $k \mapsto c_k$ becomes a function known as the Fourier transform. Denote

$L^1(\mathbb R) := \{f \in \mathcal F(\mathbb R, \mathbb C): f\ \text{is Lebesgue-integrable}\}.$

Definition 1. For any $f \in L^1(\mathbb R)$ , the Fourier transform $\hat f : \mathbb R \to \mathbb C$ is defined by

$\displaystyle \hat f(\xi) := \frac 1{2\pi} \int_{\mathbb R} f(x) e^{-i \xi x}\, \mathrm dx.$

We can extend it to a function $\hat f : \mathbb C \to \mathbb C$ as long as $\mathrm{Im}(\xi) \leq 0$ i.e. if $\xi = \sigma + i \tau$ and $\tau \leq 0$ , then the integral on the right-hand side

$\displaystyle \hat f(\sigma + i \tau) := \frac 1{2\pi} \int_{\mathbb R} f(x) \cdot e^{\tau x} e^{-i \sigma x}\, \mathrm dx$

will still converge as desired.

Remark 1. For any $s \in \mathbb C$ with $\mathrm{Re}(s) \geq 0$ , $\mathrm{Im}(-is) = -\mathrm{Re}(s) \leq 0$ , so that

$\begin{aligned} \hat f(-i s) &= \frac 1{2\pi} \int_{\mathbb R} f(t) e^{- i (-is) t}\, \mathrm dt \\ &= \frac 1{2\pi} \int_{\mathbb R} f(t) e^{-s t}\, \mathrm dt \\ &= \frac 1{2\pi} \cdot \mathcal L \{ f\}(s), \end{aligned}$

where the right-hand side denotes the Laplace transform of $f$ . More compactly,

$\mathcal L \{ f\}(s) = 2\pi \cdot \hat f(-i s),\quad \mathrm{Re}(s) \geq 0.$

In particular, if $\xi \in \mathbb R$ , then $\mathrm{Im}(\xi) = \mathrm{Re}(i\xi) = 0$ , so that the equation

$\displaystyle \hat f(\xi) = \hat f(-i(i\xi)) = \frac 1{2\pi} \cdot \mathcal L\{f\}(i \xi)$

is well-defined.

We make this connection now, so that we can rigorously define the inverse Laplace transform in terms of the inverse Fourier transform, made possible by the Fourier inversion theorem.

Depending on usage or convenience, we denote $\mathcal F \{ f \} = \hat f$ . To emphasise the underlying variables, we denote $\mathcal F\{f(x)\} = \hat f(\xi)$ .

For any subset $K \subseteq \mathbb R$ , denote $f_K = f \cdot \mathbb I_K$ .

Example 1. For piecewise continuously differentiable $2\pi$ -periodic functions $f$ , the Fourier coefficients $c_k$ of the Fourier series of $f$ are given by $c_k = \hat f_{[-\pi, \pi]}(k)$ .

Example 2. For any $a > 0$ , $\hat{\mathbb I}_{[-a,a]}$ is continuous on $\mathbb R$ such that for $\xi \neq 0$ ,

$\displaystyle \hat{\mathbb I}_{[-a, a]}(\xi) = \frac{ \sin(a \xi) }{ \pi\xi }.$

Proof. By the fundamental theorem of calculus, for $\xi \neq 0$ ,

$\begin{aligned}\hat{\mathbb I}_{[-a, a]}(\xi) &= \frac 1{2\pi} \int_{\mathbb R} \mathbb I_{[-a, a]}(x) \cdot e^{-i \xi x}\, \mathrm dx \\ &= \frac 1{2\pi} \int_{-a}^{a} e^{-i \xi x}\, \mathrm dx \\ &= \frac 1{2\pi} \cdot \frac{e^{-i a\xi} - e^{i a\xi}}{- i \xi} \\ &= \frac 1{ \pi\xi } \cdot \frac{e^{i a\xi} - e^{-i a\xi}}{ 2i} = \frac{\sin(a \xi)}{\pi \xi}. \end{aligned}$

By the usual sine limit,

$\begin{aligned} \lim_{\xi \to 0}\hat{\mathbb I}_{[-a, a]}(\xi) &= \frac {a}{\pi} \cdot \lim_{\xi \to 0} \frac{\sin(a \xi)}{a \xi} = \frac{a}{\pi} \cdot 1 \\ &= \frac{a}{\pi} = \frac 1{2\pi} \int_{-\infty}^{\infty} \mathbb I_{[-a, a]}(x) \, \mathrm dx = \hat{\mathbb I}_{[-a, a]}(0), \end{aligned}$

establishing continuity.

Example 3. $\mathcal F\{x f(x)\} = i\hat f'(\xi)$ .

Proof. Differentiating under the integral sign,

$\begin{aligned} \mathcal F\{x f(x)\} &= \frac 1{2\pi} \int_{\mathbb R} xf(x) e^{-i \xi x}\, \mathrm dx \\ &= \frac 1{2\pi} \int_{\mathbb R} f(x) \cdot i \cdot \frac{\partial}{\partial \xi} (e^{-i \xi x})\, \mathrm dx \\ &= i \cdot \frac{\mathrm d}{\mathrm d\xi} \left( \frac 1{2\pi} \int_{\mathbb R} f(x) e^{-i \xi x}\, \mathrm dx \right) = i \hat f'(\xi). \end{aligned}$

Theorem 1. For any $a > 0$ , $\displaystyle \mathcal F\{ e^{-ax^2}\} = \frac 1{2\sqrt{a\pi}} \cdot e^{-\frac{\xi^2}{4a}}$ .

Proof. We need to evaluate the integral

$\displaystyle \mathcal F\{ e^{-ax^2}\} = \frac 1{2\pi} \int_{\mathbb R} e^{-ax^2} e^{-i \xi x}\, \mathrm dx = \frac 1{2\pi} \cdot \lim_{R \to \infty} \int_{-R}^R e^{-(ax^2+i \xi x)}\, \mathrm dx.$

Completing the square,

$\displaystyle ax^2 + i\xi x = a\left( x + \frac{i \xi}{2a} \right)^2 + \frac{\xi^2}{4a},$

so that

$\displaystyle \mathcal F\{ e^{-ax^2}\} =\frac 1{2\pi} \cdot e^{-\frac{\xi^2}{4a}} \cdot \lim_{R \to \infty} \int_{-R}^R e^{-a\left( x + \frac{i \xi}{2a} \right)^2 }\, \mathrm dx.$

To evaluate the right-hand side, we return to contour integration. Define the entire function $g(z) := e^{-az^2}$ . Consider the anti-clockwise rectangle

$\Gamma : -R \to R \to R + i\xi/2a \to -R + i\xi/2a.$

Denote the corresponding by $\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4$ respectively. By the Cauchy-Goursat theorem,

$\displaystyle \oint_{\Gamma} g(z)\, \mathrm dz = 0.$

We parameterise the integrals over the four sides:

$\begin{aligned} \oint_{\Gamma_1} g(z)\, \mathrm dz &= \int_{-R}^{R} g(t)\, \mathrm dt \\ \oint_{\Gamma_2} g(z)\, \mathrm dz &= i\int_{0}^{ \xi/2a} g(R + it)\, \mathrm dt, \\ \oint_{\Gamma_3} g(z)\, \mathrm dz &= -\int_{-R}^{R} g(t + i \xi/2a)\, \mathrm dt, \\ \oint_{\Gamma_4} g(z)\, \mathrm dz &= -i\int_0^{\xi/2a} g(-R + it )\, \mathrm dt, \end{aligned}$

so that

$\displaystyle \left( \oint_{\Gamma_1} + \oint_{\Gamma_3}\right) g(z)\, \mathrm dz = - \left( \oint_{\Gamma_2} + \oint_{\Gamma_4}\right) g(z)\, \mathrm dz.$

On $\Gamma_2$ ,

$|g(R+it)| = |e^{-a(R+it)^2}| = e^{-a(R^2 - t^2)} \leq e^{\xi^2/4a^2} \cdot e^{-aR^2}.$

Similarly, on $\Gamma_4$ , $|g(-R+it)| \leq e^{\xi^2/4a^2} \cdot e^{-aR^2}$ . By the ML-inequality,

$\begin{aligned}\left| \left( \oint_{\Gamma_2} + \oint_{\Gamma_4}\right) g(z) \right| &\leq 2 \cdot e^{\xi^2/4a^2} \cdot e^{-aR^2}. \end{aligned}$

Therefore,

$\begin{aligned} \left| \oint_{\Gamma_3} g(z)\, \mathrm dz - \oint_{\Gamma_1} g(z)\, \mathrm dz \right| &= \left| \left( \oint_{\Gamma_2} + \oint_{\Gamma_4}\right) g(z)\, \mathrm dz \right| \leq 2 \cdot e^{\xi^2/4a^2} \cdot e^{-aR^2}. \end{aligned}$

Taking $R \to \infty$ , since the Gaussian integral yields

$\displaystyle \lim_{R \to \infty} \oint_{\Gamma_1} g(z)\, \mathrm dz = \lim_{R \to \infty} \int_{-R}^{R} e^{-at^2}\, \mathrm dt = \sqrt{\pi/a},$

we have

$\displaystyle \lim_{R \to \infty} \oint_{\Gamma_3} g(z)\, \mathrm dz = \lim_{R \to \infty} \oint_{\Gamma_1} g(z)\, \mathrm dz = \sqrt{\pi/a}.$

Therefore,

$\begin{aligned} \lim_{R \to \infty} \int_{-R}^{R} e^{-a(x + i \xi/2a)^2}\, \mathrm dx &= \lim_{R \to \infty} \int_{-R}^{R} g(x + i \xi/2a)\, \mathrm dx \\ &= \lim_{R \to \infty} \oint_{\Gamma_3} g(z)\, \mathrm dz = \sqrt{\pi/a}. \end{aligned}$

Finally,

$\displaystyle \mathcal F\{ e^{-ax^2}\} = \frac 1{2\pi} \cdot e^{-\frac{\xi^2}{4a}} \cdot \sqrt{\frac{\pi}{a}} = \frac 1{2\sqrt{a\pi}} \cdot e^{-\frac{\xi^2}{4a}}.$

Remark 1. Setting $a = 1/2$ , we obtain $\displaystyle \mathcal F\{ e^{-x^2/2}\} = \frac 1{\sqrt{2\pi}} \cdot e^{-\xi^2/2}$ .

We explore inverting the Fourier transform in the next post, in order to finally establish the injectivity of Laplace transforms.

—Joel Kindiak, 26 Aug 25, 1920H

January 27, 2026
Fourier Convergence

In a sense, I’m most excited to begin this segment in the discussion on complex analysis. Fourier analysis has been surprisingly fascinating for me to self-study, but has eluded my genuine study since I did not take complex analysis seriously as an undergraduate student.

To recap, for $2\pi$ -periodic functions $f : \mathbb R \to \mathbb R$ , $f = f(\cdot + 2\pi)$ that are integrable on $[-\pi, \pi]$ , we can write their Fourier series $S(f)$ in the form

$\displaystyle (S_n(f))(t) := \sum_{k=-n}^{n} \hat f(k) e^{i kt},\quad \hat f(k) := \frac 1{2\pi} \int_{-\pi}^{\pi} f(t) e^{-ikt}\, \mathrm dt,$

where the right-hand side is the integral of the complex-valued function $\mathbb R \to \mathbb C, t \mapsto f(t) e^{-ikt}$ , prompting the use of complex-analytic techniques in the future. The function $\hat f : \mathbb R \to \mathbb C$ defined by

$\displaystyle \hat f(\xi) := \frac{1}{2\pi} \int_{-\infty}^{\infty} f(t) e^{-i \xi t}\, \mathrm dt$

also turns out to be the famous Fourier transform that we will explore later on as well (where $f$ is suitably defined, of course).

However, most computational treatments assume, rather than prove, that $S_n(f) \to f$ as $n \to \infty$ via whatever meaningful mode of convergence. Having developed needful complex analytic machinery, we aim to establish said convergence under sufficiently mild conditions on $f$ . We will follow the exposition written here.

Theorem 1. Let $f : [-\pi, \pi] \to \mathbb R$ be continuously differentiable so that its periodic extension given by $f = f(\cdot + 2\pi)$ is piecewise continuously differentiable. Define $\tilde f = f$ on points of continuity and

$\displaystyle \tilde f(t) := \frac{f(t^-) + f(t^+)}{2},\quad f(t^{\pm}) := \lim_{\delta \to 0^+} f(t \pm \delta),$

whenever $t$ is a point of discontinuity (i.e. $t$ is an odd multiple of $\pi$ ). Then $\displaystyle S_n(f) \to \tilde f$ pointwise.

Theorem 1 tells us that most of the various Fourier series problems students encounter are indeed point-wise equal to the functions they represent (in fact, many of these problems are designed to be continuous even at odd multiples of $\pi$ ), so that direct substitution (e.g. $f(0) = (S_\infty(f))(0)$ ) for problem solving is mathematically valid.

Lemma 1. For any $t \in \mathbb R$ ,

$\displaystyle (S_n(f))(t) = \frac 1{\pi} \int_{-\pi}^{\pi} f(u) \cdot D_n(t-u)\, \mathrm du,$

where the function $D_n$ defined by

$\displaystyle D_n(\theta) := \frac{ \sin((n+1/2) \theta) }{ 2 \sin( \theta /2) }$

is called the Dirichlet kernel, and the integral (without the $1/\pi$ factor) is known as the convolution of $f$ and $D_n$ , sometimes abbreviated to $f * D_n$ .

Proof. By expanding the definitions of all relevant terms (here, we regard $\hat f := \hat f|_{[-\pi, \pi]}$ ),

$\begin{aligned} (S_n(f))(t) &= \sum_{k=-n}^{n} \hat f(k) e^{i kt} \\ &= \sum_{k=-n}^{n} \left( \frac{1}{2\pi} \int_{-\pi}^{\pi} f(u) e^{-i k u}\, \mathrm du \right) e^{i kt} \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(u) \left( \sum_{k=-n}^{n} e^{i k (t-u)} \right)\, \mathrm du. \end{aligned}$

To simplify the sum, we use a geometric series with common ratio $r := e^{i(t-u)}$ and first term $r^{-n}$ :

$\begin{aligned} \sum_{k=-n}^{n} e^{i k (u-t)} &= \frac {r^{-n} \cdot (1 - r^{2n+1})}{1 - r} = \frac{ r^{-(n+1/2)} - r^{n+1/2} }{ r^{-1/2}-r^{1/2} }\end{aligned}$

Denoting $\theta := t-u$ , we observe that for any $m$ , using Euler’s formula,

$\begin{aligned} r^{-m} - r^m &= e^{-i m \theta} - e^{i m \theta} = -2i \cdot \frac{ e^{i m \theta} - e^{-i m \theta} }{ 2i } = -2i \cdot \sin(m \theta). \end{aligned}$

Hence,

$\begin{aligned} \sum_{k=-n}^{n} e^{-i k \theta} = \frac{ -2i \cdot \sin((n+1/2) \theta) }{ -2i \cdot \sin( \theta /2) } = \frac{ \sin((n+1/2) \theta) }{ \sin( \theta /2) }.\end{aligned}$

Lemma 2. For any $n$ ,

$\begin{aligned}(\tilde f - S_n(f))(t) &= \frac 1{2 \pi} \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot (\sin(nv) \cot(v/2) + \cos(nv))\, \mathrm dv.\end{aligned}$

Proof. By Lemma 1, since $D_n$ is an even function,

$\begin{aligned} (S_n(f))(t) &= \frac 1{\pi} \int_{-\pi}^{\pi} f(u) \cdot D_n(t-u)\, \mathrm du \\ &= \frac 1{\pi} \int_{-\pi-t}^{\pi-t} f(t + v) \cdot D_n (-v) \, \mathrm dv \\ &= \frac 1{\pi} \int_{-\pi-t}^{\pi-t} f(t + v) \cdot D_n (v) \, \mathrm dv \\ &= \frac 1{\pi} \left( \int_{-\pi-t}^{-\pi} + \int_{-\pi}^{\pi} + \int_{\pi}^{\pi - t} \right)f(t + v) \cdot D_n (v) \, \mathrm dv \\ &= \frac 1{\pi} \left( \int_{-\pi-t}^{-\pi} + \int_{-\pi}^{\pi} - \int_{\pi-t}^{\pi } \right)f(t + v) \cdot D_n (v) \, \mathrm dv \\ &= \frac 1{\pi}\int_{-\pi}^{\pi} f(t + v) \cdot D_n (v) \, \mathrm dv \end{aligned}$

where $\int_{-\pi-t}^{-\pi} = \int_{\pi-t}^{\pi}$ by $2\pi$ -periodicity. Now

$\begin{aligned} \int_{-\pi}^{\pi} D_n(v)\, \mathrm dv &= \frac 12 \cdot \sum_{k=-n}^{n} \int_{-\pi}^{\pi} e^{-i k v}\, \mathrm dv \\ &= \frac 12 \cdot \int_{-\pi}^{\pi} e^{-i 0 v}\, \mathrm dv \\ &= \frac 12 \cdot 2\pi = \pi, \end{aligned}$

so that

$\displaystyle \frac 1{\pi} \int_{-\pi}^{\pi} \tilde f(t) \cdot D_n(v)\, \mathrm dv = \tilde f(t)$

implies that

$\begin{aligned}(\tilde f - S_n(f))(t) &= \frac 1{\pi} \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot D_n(v)\, \mathrm dv \\ &= \frac 1{2 \pi} \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot (\sin(nv) \cot(v/2) + \cos(nv))\, \mathrm dv. \end{aligned}$

Lemma 3. For $f$ continuously differentiable on $[-\pi, \pi]$ , $\hat f(\xi) \to 0$ for natural numbers $\xi \to \infty$ . This result can be further generalised into the Riemann-Lebesgue lemma.

Proof. Denote $M := \max_{t \in [-\pi, \pi]} |f'(t)|$ , which exists since $f'$ is continuous. Integrating by parts,

$\begin{aligned}|\hat f(\xi)| &= \left| \int_{-\pi}^{\pi} f(t) e^{-i \xi t}\, \mathrm dt \right| \\ &= \left| \left[ \frac{ f(t) e^{-i \xi t} }{ -i \xi } \right]_{-\pi}^{\pi} - \int_{-\pi}^{\pi} \frac{ f'(t) e^{-i \xi t} }{ -i \xi }\, \mathrm dt \right| \\ &\leq \frac 1{\xi} \left[ |f(\pi)| + | f(-\pi ) | + \int_{-\pi}^{\pi} |f'(t)|\, \mathrm dt \right] \\ &\leq \frac 1{\xi} \cdot [ |f(\pi)| + | f(-\pi ) | + 2\pi M]. \end{aligned}$

Taking $\xi \to \infty$ , $|\hat f(\xi)| \to 0$ , which implies $\hat f(\xi) \to 0$ .

Proof of Theorem 1. We pick up from Lemma 2:

$\begin{aligned}(\tilde f - S_n(f))(t) &= \frac 1{2 \pi} \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot (\sin(nv) \cot(v/2) + \cos(nv))\, \mathrm dv.\end{aligned}$

We claim that the two integrals $I_1(n), I_2(n)$ defined by

$\begin{aligned} I_1(n) &= \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot \sin(nv) \cot(v/2) \, \mathrm dv,\\ I_2(n) &= \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot \cos(nv)\, \mathrm dv,\end{aligned}$

tend to $0$ as $n \to \infty$ . For the first integral, we use the classic tangent limit

$\begin{aligned} \lim_{v \to 0} \frac{ \tilde f(t) - f(t+v) }{ \tan(v/2) } &= 2 \cdot \lim_{v \to 0} \frac{f(t) - f(t+v)}{v} \cdot \lim_{v \to 0} \frac{v/2}{\tan(v/2)} = 2 f'(t), \end{aligned}$

so that the function $g(t):= (\tilde f(t) - f(t+v)) \cdot \cot(v/2)$ is bounded, and hence piecewise continuously differentiable on $[-\pi, \pi]$ . By Lemma 3,

$\displaystyle |I_1(n)| \leq |\hat g(n)| \to 0.$

Since $h(t):= \tilde f(t) - f(t+v)$ is continuously differentiable on $[-\pi, \pi]$ , by Lemma 3, $|I_2(n)| \leq |\hat h(n)| \to 0$ .

Therefore, for any $t \in \mathbb R$

$\begin{aligned} |(\tilde f - S_n(f))(t)| &\leq |I_1(n)| + |I_2(n)| \\ &\to 0 + 0 = 0, \end{aligned}$

yielding $S_n(f) \to \tilde f$ point-wise, as required.

Couple point-wise convergence with an integrable upper-bound, we can use the dominated convergence theorem to resolve most integration-related problems. For differentiation, however, we would require the stronger technology of uniform convergence. Whether we need such technology or not, however, is up for grabs.

For now, we want to develop the notion of the Fourier transform more carefully. This we will do in the next post.

—Joel Kindiak, 25 Aug 25, 2244H

January 20, 2026
Cauchy’s Residue Theorem

We developed the notion of Laurent series in order to more easily compute integrals. In particular, the powerful technique of Cauchy’s residue theorem becomes accessible to our use.

Lemma 1. Suppose $f : \mathbb C \to \mathbb C$ is holomorphic on $\bar B(z_0, r) \backslash \{z_0\}$ with Laurent series

$\displaystyle f(z) = \sum_{k=-\infty}^\infty c_k \cdot (z - z_0)^k.$

Defining $\Gamma_0 := \partial B(z_0, r)$ with anti-clockwise direction (i.e. parameterisation $\mathbf r(t) = z_0 + re^{it}$ for $t \in [0, 2\pi]$ ),

$\displaystyle \oint_{\Gamma_0} f(z)\, \mathrm dz = 2 \pi i \cdot c_{-1}.$

Proof. By the uniqueness of Laurent coefficients,

$\displaystyle c_{-1} = \frac 1{2\pi i} \oint_{\Gamma_0} \frac{f(z)}{ (z-z_0)^{-1+1} }\, \mathrm dz = \frac 1{2\pi i} \oint_{\Gamma_0} f(z)\, \mathrm dz.$

Theorem 1 (Cauchy’s Residue Theorem). Using the notation in Lemma 1, define the residue of $f$ at $z_0$ by $\mathrm{Res}(f, z_0)$ . Suppose $f$ is holomorphic on an open set $U$ containing a simple closed contour $\Gamma = \partial D$ (directed anti-clockwise) and its interior $D$ , except for isolated singularities at $z_1,\dots, z_n$ . Then

$\displaystyle \oint_{\Gamma} f(z)\, \mathrm dz = 2\pi i \cdot \sum_{k=1}^n \mathrm{Res}(f, z_k).$

Note that if $f$ has no singularities, then we recover the Cauchy-Goursat theorem.

Proof. Fix circles $\Gamma_k := B(z_k, r)$ for sufficiently small $r > 0$ , so that the hypotheses in Lemma 1 are satisfied, and that

$\displaystyle \oint_{\Gamma_k} f(z)\, \mathrm dz = 2\pi i \cdot \mathrm{Res}(f, z_k).$

By the Cauchy-Goursat theorem and Cauchy’s integral formula,

$\begin{aligned} \oint_{\Gamma}f(z)\, \mathrm dz &= \sum_{k=1}^n \oint_{\Gamma_k} f(z)\, \mathrm dz = 2\pi i \cdot \sum_{k=1}^n \mathrm{Res}(f, z_k). \end{aligned}$

That felt a little anti-climatic, no? Well, the key trick is this: if we have simple techniques to compute the right-hand side, then the left-hand side no longer requires the painful work of re-parametrisation, and becomes much easier to compute.

Lemma 2. Suppose $f$ has a pole of order $m$ at $z_0$ . Then

$\displaystyle \mathrm{Res}(f, z_0) = \lim_{z \to z_0} \left[ \frac{\mathrm d^{n-1}}{\mathrm dz^{m-1}} ( (z-z_0)^m f(z) ) \right].$

Proof. Consider the Laurent series of $f$ given by

$\displaystyle f(z) = \sum_{k=-m}^\infty c_k (z- z_0)^k,$

since $z_0$ is a pole of order $m$ . Then

$\displaystyle (z-z_0)^m f(z) = \sum_{k = 0}^\infty c_{k-m} (z - z_0)^{k}$

has a removable singularity at $z_0$ and extends to a function $g$ holomorphic at $z_0$ . Taking $(m-1)$ derivatives,

$\displaystyle g^{(m-1)}(z) = \sum_{k = m-1}^\infty c_{k-m} \cdot \frac{k!}{(k-m+1)!} \cdot (z - z_0)^{k-(m-1)}$

Setting $z = z_0$ ,

$\begin{aligned} \lim_{z \to z_0} \frac{\mathrm d^{n-1}( (z-z_0)^m f(z) )}{\mathrm dz^{m-1}} &= g^{(m-1)}(z_0) \\ &= (m-1)! \cdot c_{-1} \\ &= (m-1)! \cdot \mathrm {Res}(f, z_0), \end{aligned}$

as required.

I’m gonna be honest—that looks even more frightening than the basic Cauchy residue theorem. True. But something beautiful happens when we deal only with simple poles.

Corollary 1. Suppose $f$ has a simple at $z_0$ . Then

$\displaystyle \mathrm{Res}(f, z_0) = \lim_{z \to z_0} [ (z-z_0) f(z) ].$

Now that looks far more pleasant. Let’s test out this new machinery with a not-so-trivial example.

Example 1. Evaluate $\displaystyle \lim_{R \to \infty} \int_{-R}^{R} \frac{\cos x + \sin x}{1 + x^2}\, \mathrm dx$ .

Solution. Recall that $\sin x + \cos x = \sqrt 2 \sin(x + \pi/4)$ , so that we want to equivalently evaluate

$\displaystyle \lim_{R \to \infty} \int_{-R}^{R} \frac{\sqrt 2 \sin(x + \pi/4)}{1 + x^2}\, \mathrm dx.$

Define the holomorphic function $f : \mathbb C \backslash \{\pm i\} \to \mathbb C$ by

$\displaystyle f(z) = \frac{e^{i z }}{1 + z^2},$

so that the desired integral is

$\begin{aligned} \displaystyle \lim_{R \to \infty} \int_{-R}^{R} \frac{\sin(x + \pi/4)}{1 + x^2}\, \mathrm dx &= \mathrm{Im} \left(\sqrt 2 e^{i \pi/4} \cdot \lim_{R \to \infty} \int_{-R}^{R} \frac{e^{iz}}{1 + z^2}\, \mathrm dz\right) \\ &= \mathrm{Im} \left(\sqrt 2 e^{i \pi/4} \cdot \lim_{R \to \infty} \int_{-R}^{R} f(z)\, \mathrm dz\right). \end{aligned}$

Consider the contour $\Gamma_R : -R \to R \to -R$ directed anti-clockwise comprising of $\Gamma_1 := [-R, R]$ and the anti-clockwise semicircle with centre $0$ and radius $R$ parameterised by $\Gamma_2 := r(t) = Re^{it}$ , $t \in [0, \pi]$ . By definition,

$\displaystyle \int_{-R}^{R} f(z)\, \mathrm dz = \oint_{\Gamma_1} f(z)\, \mathrm dz.$

For $R > 2$ , since $|z| = R$ so that

$|1 + z^2| \geq |z|^2 - 1 = R^2 - 1 > 0,$

$|e^{iz}| \leq 1$ on $\Gamma_2$ , and $\| \Gamma_2 \| = R \pi$ , by the ML-inequality,

$\displaystyle \left| \oint_{\Gamma_2} f(z)\, \mathrm dz \right| = \left| \oint_{\Gamma_2} \frac{e^{iz}}{1 + z^2}\, \mathrm dz \right| \leq \frac 1{R^2 - 1} \cdot R \pi.$

On the other hand, since the only residue of $f$ in $\Gamma$ is the simple pole $i$ , by Cauchy’s residue theorem,

$\begin{aligned} \oint_{\Gamma} f(z)\, \mathrm dz &= 2\pi i \cdot \mathrm{Res}(f, i) \\ \int_{\Gamma_1} f(z)\, \mathrm dz + \int_{\Gamma_2} f(z)\, \mathrm dz &= 2 \pi i \cdot \lim_{z \to i} [ (z-i) f(z) ] \\ \int_{-R}^{R} f(z)\, \mathrm dz + \int_{\Gamma_2} f(z)\, \mathrm dz &= 2 \pi i \cdot \lim_{z \to i} \left[ (z-i) \cdot \frac{e^{iz}}{z^2 + 1} \right] \\ &= 2 \pi i \cdot \lim_{z \to i} \left[ \frac{e^{iz}}{z + i} \right] \\ &= 2 \pi i \cdot \frac{e^{i \cdot i}}{i + i} = \frac{\pi}{e}. \end{aligned}$

Then

$\begin{aligned} \left| \int_{-R}^{R} f(z)\, \mathrm dz - \frac{\pi}{e} \right| &= \left| \int_{\Gamma_2} f(z)\, \mathrm dz \right| \leq \lim_{R \to \infty}\frac{R}{R^2 - 1}. \end{aligned}$

Taking $R \to \infty$ and using continuity,

$\displaystyle \lim_{R \to \infty } \int_{-R}^{R} \frac{e^{iz}}{z^2+1}\, \mathrm dz = \frac{\pi}{e}.$

Therefore,

$\begin{aligned}\lim_{R \to \infty} \int_{-R}^{R} \frac{\sin(x + \pi/4)}{1 + x^2}\, \mathrm dx &= \mathrm{Im} \left(\sqrt 2 e^{i \pi/4} \cdot \frac{\pi}{e} \right) \\ &= \frac{\pi}{e} \cdot \sqrt 2 \sin(\pi/4) = \frac {\pi}{e}. \end{aligned}$

A rather surprising integral! I’m convinced there’s no purely real-variable calculus method to compute this integral. The satisfying part of this example is its utility of the all of the topics we have discussed before: holomorphicity, contour integration, the ML-inequality, and Cauchy’s residue theorem that in turn required the massive complex-analytic theorems: the Cauchy-Goursat theorem, Cauchy’s integral formula, and Laurent series.

This technique is also often known as the calculus of residues, whose tools we finally have access to in studying Fourier transforms.

—Joel Kindiak, 25 Aug 25, 1445H

January 13, 2026
Badly-Behaved Points
Previously, we discussed Taylor series and their cousin Laurent series:

$\displaystyle f(z) = \sum_{k=-\infty}^{\infty} c_k (z-z_0)^k,\quad z \in \mathrm{Ann}(z_0, r_1, r_2) ,$

where

$\displaystyle c_n = \frac 1{2\pi i} \oint_{\partial B(z_0, r)} \frac{f(w)}{(w-z_0)^{n+1}}\, \mathrm dw ,\quad r_1 < r < r_2.$

The Laurent series exists (and converges uniformly) whenever $f$ is holomorphic on $\overline{\text{Ann}} (z_0, r_1, r_2)$ .

Lemma 1. Let $f : \mathbb C \to \mathbb C$ be holomorphic on $\mathrm{Ann}(z_0, r_1, r_2)$ with Laurent series

$\displaystyle f(z) = \sum_{k=-\infty}^{\infty} c_k (z-z_0)^k,\quad z \in \mathrm{Ann}(z_0, r_1, r_2).$

If $c_k = 0$ for $k < 0$ , then there exists a unique function $F : \mathbb C \to \mathbb C$ holomorphic on $B(z_0, r_2)$ such that $F|_{\mathrm{Ann}(z_0, r_1, r_2) } = f|_{\mathrm{Ann}(z_0, r_1, r_2) }$ .

Definition 1. Given the setup in Lemma 1, we call $z_0$ an isolated singularity if $f$ is holomorphic on $\bar B(z_0, r) \backslash \{z_0\}$ for some $r > 0$ . Furthermore, we call $z_0$ :
- a removable singularity if $c_k = 0$ for $k < 0$ ,
- a pole of order $m$ if $m$ is the largest integer such that $a_{-m} \neq 0$ ,
- an essential singularity if $c_k \neq 0$ for infinitely many $k < 0$ .
A pole of order $1$ (resp. $2$ ) is called a simple (resp. double) pole.

Denote $\hat{\mathbb C} := \mathbb C \cup \{ \infty \}$ for brevity.

Lemma 2. An isolated singularity $z_0$ of $f$ is removable if and only if $f$ approaches a limit in $\mathbb C$ as $z \to z_0$ .

Proof. In the direction $(\Rightarrow)$ , $f(z) \to c_0$ as $z \to z_0$ . In the direction $(\Leftarrow)$ , suppose $z_0 = 0$ without loss of generality. Since $f$ has a finite limit at $0$ , $f$ is bounded on some $B(0, r) \supseteq \bar B(0, r) \backslash \{0\}$ . Assume $|f| \leq 1$ without loss of generality. Hence, we can compute the Laurent coefficients $c_{-k}$ , $k \in \mathbb N^+$ by

$\begin{aligned}2 \pi i \cdot c_{-k} &= \oint_{\partial B(0, r)} \frac{f(w)}{w^{-k+1}}\, \mathrm dw \end{aligned}$

For any $w \in \partial B(0, r)$ , $|f(w)/w^{-k+1}| = |f(w)| / |w|^{-k+1} \leq 1 / r^{-k+1}$ , By the ML-inequality,

$\begin{aligned}2\pi \cdot |c_{-k}| = |2 \pi i \cdot c_{-k}| &\leq \frac 1{r^{-k+1}} \cdot 2 \pi r = 2 \pi r^k \quad \Rightarrow \quad |c_{-k}| \leq r^k.\end{aligned}$

Taking $r \to 0^+$ , $c_{-k} = 0$ .

Lemma 3. An isolated singularity $z_0$ of $f$ is a pole of order $m$ if and only if $(z-z_0)^m f(z)$ has a limit in $\mathbb C \backslash \{0\}$ as $z \to z_0$ .

Proof. In the direction $(\Rightarrow)$ , the Laurent series of $f$ is given by

$\displaystyle f(z) = \sum_{k=-m}^{\infty} c_k (z-z_0)^k,\quad z \in \bar B(z_0, r) \backslash \{z_0\}.$

Multiplying by $(z-z_0)^m$ ,

$\displaystyle (z-z_0)^mf(z) = \sum_{k=-m}^{\infty} c_k (z-z_0)^{m+k} = \sum_{k=0}^{\infty} c_{k-m} (z-z_0)^{k}.$

By Lemma 2, since the right-hand side has a limit $c_{-m}$ as $z \to z_0$ , so also $(z-z_0)^mf(z) \to c_{-m} \neq 0$ .

Lemma 4. An isolated singularity $z_0$ of $f$ is a pole of some order if and only if $|f(z)| \to \infty$ as $z \to z_0$ .

Proof. Suppose $z_0 = 0$ for simplicity. In the direction $(\Rightarrow)$ , $f$ has a Laurent series given by

$\displaystyle f(z) = \frac{a_{-m}}{z^m} + \sum_{k=1}^{m-1} \frac{a_{-k}}{z^m} \cdot z^{m-k} + \sum_{k=0}^{\infty} a_k z^k.$

The Taylor series portion converges to $a_0$ as $z \to 0$ . For the other portions, we use the reverse triangle inequality for sufficiently small $|z|$ to obtain $|f(z)| \to \infty$ .

For the reverse implication, suppose $|f(z)| \to \infty$ as $z \to 0$ . First, restrict $r >0$ such that $|f| |_{B(0, r) \backslash \{0\}} > 1$ . Hence, $g:=1/f$ is bounded on $B(0, r) \backslash \{0\}$ . Using the argument in Lemma 2, $0$ is a removable singularity of $g$ . Since $|f(z)| \to \infty$ , $g(z) \to 0$ as $z \to 0$ . Extend $g$ to a holomorphic function $G$ on $B(0, r) \backslash \{0\}$ , so that

$\displaystyle G(0) = \lim_{z \to 0} G(z) = \lim_{z \to 0} g(z) = 0.$

Find the largest $m$ such that $g(z) = z^m \cdot h(z)$ , where $h$ is a holomorphic function that has no root in $B(0, r)$ . Then

$\displaystyle f(z) = \frac 1{z^m} \cdot \frac 1{h(z)}.$

Since $1/h$ has no root in $B(0, r)$ , we can expand it into its Taylor series, so that

$\displaystyle f(z) = \frac 1{z^m} \sum_{k=0}^\infty d_k z^k,\quad d_0 \neq 0.$

In particular, $f$ has a pole of order $m$ .

Lemma 5. An isolated singularity $z_0$ of $f$ is an essential singularity if and only if $f$ has no limit in $\hat{\mathbb C}$ as $z \to z_0$ .

Proof. Lemmas 2, 3, and 4.

Essential singularities are therefore relatively bad-behaved points.

Theorem 1 (Casorati-Weierstrass Theorem). Let $z_0$ be an essential singularity of $f$ . Then for any $\delta > 0, w \in \mathbb C, \epsilon > 0$ ,

$f(B(z_0, \delta)\backslash \{z_0\}) \cap B(w,\epsilon) \neq \emptyset.$

Proof. Assume $z_0 = 0$ for simplicity. Suppose for a contradiction there exists $\delta, w, \epsilon$ such that the intersection is empty. Then for any $z \in B(0, \delta) \backslash \{0\}$ , $|f(z) - w| \geq \epsilon > 0$ . Define $g$ by $g(z) := 1/(f(z) - w)$ , which is bounded above by $1/\epsilon$ , and so has a removable singularity at $0$ . Find the largest integer $m$ such that $g(z) = z^m \cdot h(z)$ for some holomorphic $h$ that does not have a root in $B(0,\delta)$ . Therefore,

$\displaystyle f(z) = w + \frac 1{z^m} \cdot \frac 1{h(z)}.$

If $m =0$ , then $0$ is a removable singularity. If $m > 0$ , then $0$ is a pole of order $m$ . Either outcome yields a contradiction.

Having discussed singularities, we are now primed to talk about residues.

—Joel Kindiak, 22 Aug 25, 1458H
November 26, 2025
Infinite Series Revisited

Previously, we have seen that if $f$ is holomorphic on a domain $D$ , then all of its derivatives are holomorphic on $D$ .

Theorem 1. If $f$ is holomorphic on $D$ , then for any $z_0 \in D$ , there exists $r > 0$ such that

$\displaystyle f(z_0 + w) = \sum_{k=0}^\infty \frac{ f^{(k)}(z_0) }{ k! } \cdot w^k,\quad |w| < r.$

Furthermore, the convergence holds uniformly, and the coefficients in the series expansion are unique. The converse holds trivially.

Proof. Denoting $\Gamma := \partial B(0,r) = \{\omega \in \mathbb C : |\omega| = r\}$ for $r$ to be determined, applying Cauchy’s integral formula to the function $f(z_0 + \cdot)$ , for any $|w| < r$ , since $|w/\omega| < 1$ ,

$\begin{aligned} 2\pi i \cdot f(z_0 + w) &= \oint_{\Gamma} \frac{f(z_0 + \omega)}{\omega - w}\, \mathrm d\omega \\ &= \oint_{\Gamma} \frac{f(z_0 + w)}{\omega} \cdot \frac{1}{1 - \frac{ w }{ \omega }}\, \mathrm d\omega \\ &= \oint_{\Gamma} \frac{f(z_0 + w)}{\omega} \cdot \sum_{k=0}^\infty \left( \frac{w}{\omega} \right)^k\, \mathrm d\omega \\ &= \sum_{k=0}^\infty \oint_{\Gamma} \frac{f(z_0 + w)}{\omega^{k+1}}\, \mathrm d\omega \cdot w^k \\ &= 2\pi i \cdot \sum_{k=0}^\infty \frac{ f^{(k)}(z_0) }{ k! } \cdot w^k, \end{aligned}$

where the interchange of the infinite sum and the integral is justified since the geometric series converges uniformly. For the uniqueness claim, assume

$\displaystyle f(z_0 + w) = \sum_{k=0}^\infty c_k \cdot w^k = \sum_{k=0}^\infty d_k \cdot w^k.$

We claim that $c_n = 0$ for any $n$ . Setting $w = 0$ , $c_0 = d_0$ . Suppose $c_k = d_k$ for $k = 0,1,\dots, n$ . Differentiating $n+1$ times, uniform convergence yields

$\displaystyle (n+1)! \cdot c_{n+1} + O(w) = (n+1)! \cdot d_{n+1} + O(w).$

Setting $w = 0$ , $c_{n+1} = d_{n+1}$ , as required.

Definition 1 (Taylor Series). A function $f : \mathbb C \to \mathbb C$ is analytic at a point $z_0 \in \mathbb C$ if all of its derivatives are holomorphic at $z_0$ and there exists $r > 0$ such that

$\displaystyle f(z_0 + w) = \sum_{k=0}^\infty \frac{ f^{(k)}(z_0) }{ k! } \cdot w^k,\quad |w| < r.$

The function $f$ is analytic on a domain $D$ if it is analytic at every point in $D$ . We call the right-hand side the Maclaurin series of the function $f(z_0 + \cdot)$ . The Taylor series at $z_0$ is obtained via the substitution $z := z_0 + w$ :

$\displaystyle f(z) = \sum_{k=0}^\infty \frac{ f^{(k)}(z_0) }{ k! } \cdot (z-z_0)^k,\quad z \in B(z_0, r).$

Corollary 1. A function $f : \mathbb C \to \mathbb C$ is holomorphic on a domain $D$ if and only if it is analytic on $D$ .

When are two holomorphic functions identical to each other? Some sufficient conditions are surprisingly mild.

Theorem 2 (Identity Theorem). Let $f, g : \mathbb C \to \mathbb C$ be holomorphic on a domain $D$ . Suppose for any $z_0 \in D$ such that there exists a sequence $z_k \to z_0$ , $f(z_k) = g(z_k)$ for any $k$ . Then $f|_D = g|_D$ .

Proof. Assume without loss of generality that $g = 0$ , so that we apply the result to $f=g \iff f-g = 0$ . Fix $z_0 \in \mathbb C$ and $z_n \to z_0$ such that $f(z_k) = 0$ . By continuity, $f(z_0) = 0$ , so that $z_0$ is a root of $f$ . Fix $r> 0$ such that

$\displaystyle f(z) = \sum_{m=0}^\infty \frac{f^{(m)}(z_0)}{m!} \cdot (z-z_0)^m,\quad z \in B(z_0, r) =: B.$

Denote $c_m := f^{(m)}(z_0)/m!$ for brevity. Suppose for a contradiction that $f|_B \neq 0$ . Then there exists $M \in \mathbb N$ such that $c_M \neq 0$ . By Corollary 1, there exists some $g(z) \in O(z)$ holomorphic on $D$ such that

$\displaystyle f(z) = c_M \cdot (z - z_0)^M \cdot (1 + g(z-z_0)).$

For sufficiently large $k$ ,

$0 = f(w_k) = (w_k - z_0)^M \cdot (1 + g(w_k - z_0)) \neq 0,$

a contradiction. Therefore, $f|_B = 0$ .

Now we prove that $f|_D = 0$ . Define

$\displaystyle U := \bigcup_{S \in \mathcal U} S,\quad \mathcal U := \{B \subseteq D\ \text{open} : f|_B = 0 \}.$

Then $U$ is nonempty, open, and closed. This means that $V := D \backslash U$ is open and closed as well. Since $D$ is connected, $U \sqcup V = D$ and $U \neq \emptyset$ implies $V = \emptyset$ , so that $D = U$ . In particular, $f|_D = 0$ , as required.

Definition 2. Define the annulus with centre $z_0$ and radii $0 < r_1 < r_2$ by

$\text{Ann}(z_0, r_1, r_2) := \{ z \in \mathbb C : r_1 < |z - z_0| < r_2 \}.$

Define the closed disk by $D(z_0, r) := \bar{B}(z_0, r)$ , and the punctured disk by $D(z_0, r) \backslash \{z_0\} = \text{Ann}(z_0, 0, r)$ .

Theorem 3 (Laurent Series). Suppose $f : \mathbb C \to \mathbb C$ is holomorphic on an open set containing $\overline{\text{Ann}}(z_0, r_1, r_2)$ . Define $\Gamma_1 := \partial B(z_0, r_1)$ and $\Gamma_2 := \partial B(z_0, r_2)$ , both oriented anti-clockwise, so that $\Gamma_1 \cup \Gamma_2 = \partial \overline{\text{Ann}}(z_0, r_1, r_2)$ . Then for any $r_1<r<r_2$ ,

$\displaystyle f(z) = \sum_{k=-\infty}^{\infty} c_k (z - z_0)^k,\quad z \in \text{Ann}(z_0, r_1, r_2),$

where the convergence is uniform,

$\displaystyle c_k = \frac 1{2\pi i} \oint_{\partial B(z_0, r)} \frac{f(w)}{(w-z_0)^{k+1}}\, \mathrm dw,$

and $\partial B(z_0, r)$ is oriented anti-clockwise.

Proof. Assume $z_0 = 0$ for simplicity. By the Cauchy-Goursat theorem, for aby $z \in \text{Ann}(z_0, r_1, r_2)$ and anti-clockwise circle $\gamma \subseteq \text{Ann}(z_0, r_1, r_2)$ with centre $z$ ,

$\displaystyle \left( \oint_{\Gamma_2} - \oint_{\Gamma_1} - \oint_{\gamma} \right) \frac{ f(w) }{ w - z }\, \mathrm dw = \oint_{\Gamma_2 + (- \Gamma_1) + (-\gamma)} \frac{ f(w) }{ w - z }\, \mathrm dw = 0.$

By Cauchy’s integral formula,

$\displaystyle f(z) = \frac{1}{2\pi i} \oint_{\gamma} \frac{ f(w) }{ w - z }\, \mathrm dw = \frac{1}{2\pi i} \oint_{\Gamma_2}\frac{ f(w) }{ w - z }\, \mathrm dw - \frac{1}{2\pi i} \oint_{\Gamma_1}\frac{ f(w) }{ w - z }\, \mathrm dw.$

For the first integral, for any $w \in \Gamma_2$ , $|w| > |z| \Rightarrow |z/w| < 1$ . By the uniform convergence of the geometric series,

$\displaystyle \frac 1{w-z} = \frac 1w \cdot \frac 1{1- \frac zw} = \frac 1w \cdot \sum_{k=0}^\infty \left( \frac zw \right)^k = \sum_{k=0}^\infty \frac{1}{w^{k+1}} \cdot z^k.$

Thanks to uniform convergence,

$\displaystyle \frac{1}{2\pi i} \oint_{\Gamma_2}\frac{ f(w) }{ w - z }\, \mathrm dw = \sum_{k=0}^\infty \left( \frac{1}{2\pi i} \oint_{\Gamma_2} \frac{f(w)}{w^{k+1}}\, \mathrm dw \right) \cdot z^k.$

Similarly for the second integral, for any $w \in \Gamma_1$ , $|w| < |z| \Rightarrow |w/z| < 1$ . By the uniform convergence of the geometric series,

$\displaystyle \frac 1{w-z} = -\frac 1z \cdot \frac 1{1-\frac wz} = -\sum_{k=0}^\infty \frac{1}{w^{-k}} \cdot z^{-(k+1)} = -\sum_{k=-\infty}^{-1} \frac 1{w^{k+1}} \cdot z^{k}.$

Thanks to uniform convergence,

$\displaystyle -\frac{1}{2\pi i} \oint_{\Gamma_1}\frac{ f(w) }{ w - z }\, \mathrm dw = \sum_{k=-\infty}^{-1} \frac{1}{2\pi i} \left( \oint_{\Gamma_1} \frac {f(w)}{w^{k+1}}\, \mathrm dw \right) \cdot z^{k}.$

By the Cauchy-Goursat theorem,

$\displaystyle f(z) = \sum_{k=-\infty}^{\infty} c_k z^k,\quad c_k = \frac 1{2\pi i} \oint_{\partial B(z_0, r)} \frac{f(w)}{w^{k+1}}\, \mathrm dw.$

For uniqueness, write

$\displaystyle f(z) = \sum_{k=-\infty}^{\infty} d_k z^k.$

By the definition of $c_n$ and uniform convergence that justifies integral-summation swapping,

$\begin{aligned} 2\pi i \cdot c_n &= \oint_{\partial B(z_0, r)} \frac{f(w)}{w^{k+1}}\, \mathrm dw \\ &= \oint_{\partial B(z_0, r)} \frac{1}{w^{n+1}}\, \sum_{k=-\infty}^{\infty} d_k w^k \cdot \mathrm dw \\ &= \sum_{k=-\infty}^{\infty} d_k \oint_{\partial B(z_0, r)} \frac{1}{w^{(n-k)+1} }\, \mathrm dw \\ &= \sum_{k=-\infty}^{\infty} d_k \cdot 2\pi i \cdot \mathbb I_{n}(k) = 2 \pi i \cdot d_n.\end{aligned}$

What’s the goal of the Laurent series? Roughly speaking, the Laurent series of a function $f$ tells us how “badly behaved” it is at a point $z_0$ . If, for instance, $c_k = 0$ for $k < 0$ , then we get the usual Taylor series which tells us that $f$ is holomorphic at $z_0$ . Otherwise, even though $f$ may not be holomorphic $z_0$ , we might be able to multiply sufficiently many copies of $(z-z_0)$ so that the resulting function is holomorphic at $z_0$ . In this case, though badly behaved, the bad behavior $f$ is somewhat controllable. We will explore these singularities the next post, and develop the calculus of residues that help us compute many otherwise intractable integrals.

—Joel Kindiak, 21 Aug 25, 2236H

November 19, 2025
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

November 12, 2025
Cauchy-Goursat Theorem

Previously, we have seen that if $f$ has an antiderivative $F$ on a domain $D$ , then for any closed smooth contour $\gamma \subseteq D$ ,

$\displaystyle \oint_{\gamma} f(z)\, \mathrm dz = 0,$

and vice versa. When does this hold?

This post will take heavy reference from this document.

Theorem 1 (Cauchy-Goursat Theorem). Let $U \subseteq \mathbb C$ be open and $f : \mathbb C \to \mathbb C$ . If $f$ is holomorphic on $U$ , then for any bounded domain $D$ with $\bar D \subseteq U$ ,

$\displaystyle \oint_{\partial D} f(z)\, \mathrm dz = 0.$

Lemma 1. Theorem 1 holds whenever $\bar D$ is a closed rectangle.

Proof. Firstly, since $f$ is holomorphic on $U$ , it is continuous on $\bar D$ , so that there exists $\alpha > 0$ such that

$\displaystyle \left| \oint_{\partial D} f(z)\, \mathrm dz \right| \leq \alpha.$

Suppose for a contradiction that equality holds. Then divide $D$ into $4$ equal sub-rectangles. By the triangle inequality and the pigeonhole principle, at least one of these rectangles $D^{(1)}$ with $\| \partial D^{(1)} \| = \| \partial D \| / 2$ must yield

$\displaystyle \left| \int_{\partial D^{(1)}} f(z)\, \mathrm dz \right| \geq \frac \alpha 4.$

Inductively obtain a decreasing sequence $D =: D^{(0)} \supseteq D^{(1)} \supseteq D^{(2)} \supseteq \dots$ of rectangles such that

$\displaystyle \left| \oint_{\partial D^{(n)}} f(z)\, \mathrm dz \right| \geq \frac \alpha{4^n}.$

Each rectangle $D^{(m)}$ for each $m > 0$ has a center $z_m$ . Since the sequence $\{z_m\}$ is Cauchy, it converges to some $z_0 \in \bar D$ . Define the map $E$ by

$E(z) := f(z) - (f(z_0) + f'(z_0) \cdot (z - z_0)).$

Fix $\epsilon > 0$ . Since $f$ is holomorphic at $z_0$ , there exists $\delta > 0$ such that

$0 < |z - z_0| < \delta \quad \Rightarrow \quad |E(z)| < \epsilon.$

Since $z_m \to z_0$ , find $m$ such that $D^{(m)} \subseteq B(z_0, \delta)$ . Since the map

$f(z) - E(z) = f(z_0) + f'(z_0) \cdot (z-z_0)$

has an anti-derivative $f(z_0) \cdot (z-z_0) + f'(z_0) \cdot (z-z_0)^2/2$ ,

$\displaystyle \oint_{ \partial D^{(m)} } E(z)\, \mathrm dz = \oint_{ \partial D^{(m)} } f(z)\, \mathrm dz.$

By construction, $\|\partial D^{(m)}\| = \| \partial D \|/2^m$ . By the ML-inequality,

$\displaystyle \frac{\alpha}{4^m} \leq \left| \oint_{ \partial D^{(m)} } E(z)\, \mathrm dz \right| \leq \epsilon \cdot \frac{\| \partial D \| }{ 2^m }.$

Thus, the inequality yields a contradiction if

$\displaystyle \frac{\alpha}{2^m \cdot \| \partial D \| } > \epsilon.$

Therefore, set $\epsilon = 1$ and take $m$ to be sufficient large to produce the desired contradiction. Therefore,

$\displaystyle \left| \oint_{\partial D} f(z)\, \mathrm dz \right| = 0\quad \Rightarrow \quad \oint_{\partial D} f(z)\, \mathrm dz = 0.$

Lemma 2. Suppose that $U \subseteq \mathbb C$ is simply-connected and $f$ is holomorphic on $U$ . Then $f$ has an anti-derivative $F$ on $U$ . Equivalently,

$\displaystyle \oint_{\Gamma} f(z)\, \mathrm dz$

for any closed contour $\Gamma \subseteq D$ .

Proof. Fix $z_0 \in U$ . For any $z \in U$ , let $\Gamma_1, \Gamma_2$ be two paths from $z_0$ to $z$ comprised only of finitely many horizontal and vertical paths. Then it is not hard to prove that there exist disjoint open rectangles $D_m$ such that $\bar D_m \subseteq U$ and

$\displaystyle \partial \left( \bigsqcup_{m=1} D_m \right) = \Gamma_1 + (-\Gamma_2).$

Therefore,

$\displaystyle \begin{aligned} \oint_{\Gamma_1} f(z)\, \mathrm dz - \oint_{\Gamma_2} f(z)\, \mathrm dz &= \oint_{\Gamma_1 + (-\Gamma_2)} f(z)\, \mathrm d z \\ &= \sum_{m=1}^n \oint_{\partial D_m} f(z)\, \mathrm dz \\ &= \sum_{m=1}^n 0 = 0. \end{aligned}$

Thus, the map $F : \mathbb C \to \mathbb C$ defined by

$\displaystyle F(z) = \oint_{\Gamma} f(z)\, \mathrm dz,$

where $\Gamma$ is any path from $z_0$ to $z$ comprised only of finitely many horizontal and vertical paths is well-defined. To establish the antiderivative property, work with the path $\Gamma = \gamma_1 + \gamma_2$ where each path $\gamma_1 = r_1([0, 1])$ and $\gamma_2 = r_2([0, 1])$ is (without loss of generality) parameterised by

$r_1(t) = z_0 + t \cdot \mathrm{Re}(z - z_0),\quad r_2(t) = r_1(1) + it \cdot \mathrm{Im}(z - z_0).$

Of course, we verify that

$\begin{aligned} r_2(1) &= r_1(1) + i \cdot \mathrm{Im}(z - z_0) \\ &= z_0 + \mathrm{Re}(z - z_0) + i \cdot \mathrm{Im}(z - z_0) \\ &= z_0 + (z - z_0) = z. \end{aligned}$

so that $\Gamma$ is indeed a path from $z_0$ to $z$ . Taking $|z - z_0|$ sufficiently small and employing the continuity of $f$ at $z_0$ , the ML-inequality yields

$\displaystyle |F(z) - F(z_0) - f(z_0) \cdot (z-z_0)| = \left| \oint_{\Gamma} (f(z) - f(z_0)) \right| \leq \epsilon \cdot |z - z_0|,$

establishing differentiability at $z_0$ . Since $U$ is open and $z_0 \in U$ is arbitrary, $F$ is holomorphic on $U$ with $F' = f$ .

We now handle the general case of Theorem 1.

Proof of Theorem 1. First suppose the special case that $U$ is simply connected. Then $\partial D \subseteq U$ is a closed contour, so that by Lemma 2,

$\displaystyle \oint_{\partial D} f(z)\, \mathrm dz = 0.$

For the general case, find a finite number of line segments $\gamma_m$ with end-points in the smooth parts of $\partial D$ so that $D_0 := D \backslash \bigcup_{m=1}^n \gamma_m$ is simply connected. Intuitively, we “chop” $U$ up into a collection of simply connected “islands”. For each $\epsilon > 0$ , define

$\displaystyle \gamma_{m,\epsilon} := \left\{z \in \mathbb C : \sup_{w \in \gamma_m} | z - w | \leq \epsilon \right\},\quad D_\epsilon := D \backslash \bigcup_{m=1}^n \gamma_{m, \epsilon} \subseteq D_0.$

For sufficiently small $\epsilon$ , $\bar D_\epsilon \subseteq U$ for some simply connected neighborhood $U$ , so that

$\displaystyle \oint_{\partial D_\epsilon} f(z)\, \mathrm dz = 0.$

Then using arguments involving uniform continuity, the ML-inequality, and the definition of the contour integral, take $\epsilon \to 0^+$ to obtain

$\displaystyle \oint_{\partial D} f(z)\, \mathrm dz = \lim_{\epsilon \to 0^+} \oint_{\partial D_\epsilon} f(z)\, \mathrm dz = 0.$

We took great pains to prove the Cauchy-Goursat theorem so that we can prove an even more incredible result—the Cauchy integral formula. This we will do the next time, as well as establish the many powerful properties of holomorphic functions that set it apart in the complex analytic setting from usual differentiability in the real-analytic setting.

—Joel Kindiak, 18 Aug 25, 1756H

November 5, 2025
The Complex Integral
Let’s discuss integration in the complex world. We will use Lebesgue-integration in formulating our definitions, since we have painstakingly developed that technology elsewhere in this blog already, and can therefore leverage its many useful convergence properties, especially the Fubini–Tonelli theorem.

Definition 1. Let $f = u + iv: [a, b] \to \mathbb C$ , where $u, v :[a,b] \to \mathbb R$ . Equip $[a, b] \subseteq \mathbb R$ with the usual Lebesgue measure. If $u, v$ are Lebesgue-integrable, then we define

$\displaystyle \int_{[a, b]} f \, \mathrm d \lambda := \int_{[a, b]} u \, \mathrm d \lambda + i \int_{[a, b]} v \, \mathrm d \lambda.$

To be explicit with the variables, we make the notation $\mathrm d\lambda(t) = \mathrm dt$ to obtain

$\displaystyle \int_{[a, b]} f(t) \, \mathrm dt := \int_{[a, b]} u(t) \, \mathrm d t + i \int_{[a, b]} v(t) \, \mathrm d t.$

In the case $u, v$ are Riemann-integrable,

$\displaystyle \int_a^b f(t) \, \mathrm dt := \int_a^b u(t) \, \mathrm d t + i \int_a^b v(t) \, \mathrm d t.$

Occasionally, when there is no ambiguity, we even suppress the dummy variables for brevity:

$\displaystyle \int_a^b f = \int_a^b u + i \int_a^b v.$

Note that $u,v$ are Lebesgue-integrable if they are Riemann-integrable, and that their integrals coincide.

Lemma 1. For $f : [a, b] \to \mathbb C$ , whenever defined, $\displaystyle \left| \int_a^b f \right| \leq \int_a^b |f|$ .

Proof. First, define

$\displaystyle \theta := \mathrm{Arg} \int_a^b f\quad \Rightarrow \quad e^{-i\theta} \int_a^b f \in \mathbb R.$

Since $|e^{-i \theta}| = 1$ ,

$\begin{aligned}\left| \int_a^b f \right| &= |e^{-i\theta}| \cdot \left| \int_a^b f \right| \\ &= \left| e^{-i\theta} \int_a^b f \right| = \left| \mathrm{Re} \left( e^{-i\theta} \int_a^b f \right) \right| \\ &= \left| \int_a^b \mathrm{Re} ( e^{-i\theta} f ) \right| \leq \int_a^b \left| \mathrm{Re} ( e^{-i\theta} f ) \right| \leq \int_a^b| e^{-i\theta} f | = \int_a^b |f|.\end{aligned}$

That sounds easy enough! What about complex integration $f : \mathbb C \to \mathbb C$ ? Our idea will require the use of curves.

Definition 2. A subset $\gamma \subseteq \mathbb C$ is a curve if there exists a continuous function $r : [a, b] \to \mathbb C$ such that $\gamma = r([a, b])$ . Furthermore, we say that $\gamma$ is smooth if $r$ is continuously differentiable.

Our first line of business would be the compute the length $\| \gamma \|$ of $\gamma$ . To that end, let $P = \{ t_0, t_1,\dots, t_n\}$ be a partition of $[a, b]$ with $( t_0, t_n ) = (a, b)$ . Approximate $\|\gamma \|$ using straight-lines:

$\displaystyle L_n(\gamma) := \sum_{k=1}^n | r(t_{k+1}) - r(t_k) | = \sum_{k=1}^n \frac{ | r(t_{k+1}) - r(t_k) | }{\Delta t_k} \cdot \Delta t_k.$

Taking $n \to \infty$ so that $\Delta t_k \to 0$ (we can formalise this idea using Darboux integrals), we obtain

$\displaystyle \lim_{n \to \infty} L_n(\gamma) = \int_a^b \left| \lim_{\Delta t \to 0} \frac{ r(t + \Delta t) - r(t) }{\Delta t} \right| \mathrm dt = \int_a^b | r'(t) |\, \mathrm dt.$

Definition 3. Define the length of a smooth curve $\gamma = r([a,b])$ by

$\displaystyle \| \gamma \| := \int_a^b | r' |.$

Example 1. Let $C := r([0, 2\pi])$ where $r(t) = e^{it}$ . Then

$| r'(t) | = | i e^{ i t} | = | i | \cdot | e^{2\pi it} | = 1$

implies that

$\begin{aligned} \| C \| = \int_0^{2\pi} | r'(t) |\, \mathrm dt &= \int_0^{2\pi} 1 = 2\pi - 0 = 2\pi, \end{aligned}$

which intuitively agrees with the circumference of a unit circle equaling $2\pi$ .

To give the expression $\displaystyle \int f(z)\, \mathrm dz$ a useful meaning depends on what values of $z$ we are effectively doing an infinite summation over. If $z \in \gamma = r([a, b])$ , then $z = r(t)$ , which implies at least informally that $\mathrm dz = r'(t)\, \mathrm dt$ . This is how we define the complex integral over $\gamma$ , and establish what goes wrong if we don’t do it this way.

For a sanity check, suppose $\gamma = r_1([a, b])$ . Let $\varphi : [a, b] \to [c, d]$ denote the canonical homemorphism. Suppose $\gamma = r_2([c, d])$ where $r_2 = \varphi \circ r_1$ . Using the substitution $s = \varphi(t)$ so that $s(a) = c$ and $s(b) = d$ ,

$\begin{aligned} \int_a^b (f \circ r_1) \cdot r_1' &= \int_a^b f(r_1(t)) \cdot r_1'(t)\, \mathrm dt \\ &= \int_a^b f(r_2(\varphi (t) ))\cdot r_2'(\varphi(t)) \cdot \varphi'(t)\, \mathrm dt \\ &= \int_c^d f(r_2( s ))\cdot r_2'( s )\, \mathrm ds \\ &= \int_c^d (f \circ r_2) \cdot r_2' \end{aligned}$

Definition 4. The contour integral of $f : \mathbb C \to \mathbb C$ over $\gamma = r([a, b])$ is defined by

$\displaystyle \oint_{\gamma} f(z)\, \mathrm dz := \int_a^b f(r(t)) \cdot r'(t)\, \mathrm dt \equiv \int_a^b (f \circ r) \cdot r',$

whenever the right-hand side exists. For instance, this definition makes sense when $f$ is continuous.

Example 2. Using the circle $C$ in Example 1,

$\displaystyle \oint_C \frac 1{z}\, \mathrm dz = \int_0^{2\pi} \frac 1{e^{it}} \cdot i \cdot e^{it}\, \mathrm dt = \int_0^{2\pi} i\, \mathrm dt = 2 \pi i.$

Example 2 will play a crucial role later on when discussing Cauchy’s integral formula.

Why not just do it the usual calculus way? That is, given $f$ , compute an antiderivative $F$ such that $F' = f$ , then use the fundamental theorem of calculus to compute

$\displaystyle \int_a^b f = F(b) - F(a).$

Define the upper-half semicircle by $C^+ := r([0, \pi])$ and the lower-half semicircle by $C^- := \tilde r([0, \pi])$ , where $\tilde r := r(-\cdot)$ . Then

$\displaystyle \oint_{C^+} \frac 1z\, \mathrm dz = \pi i,\quad \oint_{C^-} \frac 1z\, \mathrm dz = -\pi i.$

However, using the endpoints of the curves $r(0) = 1$ and $r(\pi) = -1$ and the calculation $\mathrm{Log}'(z) = 1/z$ ,

$\displaystyle \pi i = \mathrm{Log}(-1) - \mathrm{Log}(1) = \int_1^{-1} \frac 1z\, \mathrm dz = \oint_{C^-} \frac 1z\, \mathrm dz = - \pi i,$

a blatant contradiction. Therefore, these definitions change depending on the paths taken between the endpoints.

Remark 1. If, instead, we took $\mathrm{Log}(-1) := \lim_{t \to \pi} \mathrm{Log}(e^{-it})$ , then we would get

$\mathrm{Log}(-1) - \mathrm{Log}(1) = -i \pi.$

Nevertheless, these various paths are not necessarily unrelated to one another.

Lemma 2. Given a curve $\gamma = r ( [a, b] )$ , define the reverse $-\gamma$ of $\gamma$ by $-\gamma = \tilde r([a, b])$ , where $\tilde r = r(a+b- \cdot)$ . For suitable $f : \mathbb C \to \mathbb C$ ,

$\displaystyle \oint_{-\gamma} f(z)\, \mathrm dz = -\oint_{\gamma} f(z)\, \mathrm dz.$

Proof. By Definition 4, making the change-of-variables $s = \tilde r$ so that $s(a) = b, s(b)= a$ ,

$\begin{aligned} \oint_{-\gamma} f(z)\, \mathrm dz &= \int_a^b (f \circ \tilde r) \cdot \tilde r' \\ &= \int_b^a (f \circ s) \cdot s' \cdot (-1) \\ &= - \int_a^b (f \circ s) \cdot s' = -\oint_{\gamma}f(z)\, \mathrm dz. \end{aligned}$

Also rather intuitively, it is not hard to integrate over curves consisting of “sub-curves”: given $\gamma_1 = r_1([a, b])$ and $\gamma_2 = r_2([b, c])$ with $r_1(b) = r_2(b)$ , define $r = r_1 \cdot \mathbb I_{[a, b]} + r_2 \cdot \mathbb I_{[b, c]}$ and $\gamma_1 + \gamma_2 := r_3([a, c])$ , so that

$\displaystyle \oint_{\gamma_1 + \gamma_2} f(z)\, \mathrm dz = \oint_{\gamma_1} f(z)\, \mathrm dz + \oint_{\gamma_2} f(z)\, \mathrm dz.$

Combining these notions leads to our first powerful lemma in complex analysis: the ML-inequality. We call $\gamma$ a contour if there exist smooth curves $\gamma_k$ such that $\Gamma = \sum_{k=1}^n \gamma_k$ , and suitably obtain

$\displaystyle \| \Gamma \| := \sum_{k=1}^n \| \gamma_k \|,\quad \oint_{\Gamma} f(z)\, \mathrm dz = \sum_{k=1}^n \oint_{\gamma_k} f(z)\, \mathrm dz.$

Henceforth, we will let $\Gamma$ denote a general contour that is the sum of possibly more than one smooth curve $\gamma$ .

Theorem 1 (ML-Inequality). Given continuous $f : \mathbb C \to \mathbb C$ and a contour $\Gamma$ with $M := \max_{z \in \Gamma} |f(z)|$ and $L := \| \Gamma \|$ ,

$\displaystyle \left| \oint_{\Gamma} f(z)\, \mathrm dz \right| \leq M \cdot L.$

Informally, the ML-inequality is the complex version of the triangle inequality that therefore becomes an exceedingly useful bounding technique in proving subsequent theorems in complex analysis.

Proof. First suppose $\gamma$ is a smooth curve. By Lemma 1,

$\begin{aligned} \left| \oint_{\gamma} f(z)\, \mathrm dz \right| &= \left| \int_a^b f(r(t)) \cdot r'(t)\, \mathrm dt \right| \\ &\leq \int_a^b |f(r(t))| \cdot |r'(t)|\, \mathrm dt \\ &\leq \int_a^b M \cdot |r'(t)|\, \mathrm dt \\ &= M \cdot \int_a^b |r'(t)|\, \mathrm dt = M \cdot L. \end{aligned}$

For the general case, denote $M := \max\{M_1,\dots, M_n\}$ and $L := \sum_{k=1}^n L_k$ . Then

$\begin{aligned} \left| \oint_{\Gamma} f(z)\, \mathrm dz \right| &= \left| \sum_{k=1}^n \oint_{\gamma_k} f(z)\, \mathrm dz \right| \\ &\leq \sum_{k=1}^n \left| \oint_{\gamma_k} f(z)\, \mathrm dz \right| \\ &\leq \sum_{k=1}^n M \cdot L_k = M \cdot \sum_{k=1}^n L_k = M \cdot L. \end{aligned}$

Let’s also wrap things up by investigating the situations when complex integration resembles real-variable integration, in terms of the fundamental theorem of calculus.

Definition 5. Let $D \subseteq \mathbb C$ be a domain and $f : D \to \mathbb C$ be continuous.
- The function $F : D \to \mathbb C$ an antiderivative of $f$ on $D$ if $F' = f$ .
- A contour $\Gamma = r([a,b])$ is closed if $r(a) = r(b)$ .
Furthermore, we call the contour integral from $z_1$ to $z_2$ is path-independent if for any $\gamma_1,\gamma_2$ with starting point $z_1$ and ending point $z_2$ ,

$\displaystyle \oint_{\gamma_1} f(z)\, \mathrm dz = \oint_{\gamma_2} f(z)\, \mathrm dz.$

Lemma 3. Let $f : [a, b] \to \mathbb C$ be continuous. If $F : [a, b] \to \mathbb C$ is an antiderivative of $f$ , then

$\displaystyle \int_a^b f' = F(b) - F(a).$

Theorem 2. Let $f : D \to \mathbb C$ be continuous. The following are equivalent:
- $f$ has an antiderivative on $D$ .
- $\oint_{\Gamma} f(z)\, \mathrm dz = 0$ for any closed contour $\Gamma \subseteq D$ .
- For any $z_1,z_2 \in D$ , the contour integral from $z_1$ to $z_2$ is path-independent.
Proof. First, assume that $f$ has an antiderivative $F$ on $D$ . By the chain rule, $(F \circ r)' = (F' \circ r) \cdot r' = (f \circ r) \cdot r'$ . Fix any contour $\gamma = r([a, b])$ from $z_1$ to $z_2$ that is smooth without loss of generality. By Lemma 3,

$\displaystyle \begin{aligned} \oint_{\gamma} f(z)\, \mathrm dz &= \int_a^b (f \circ r) \cdot r' = \int_a^b (F \circ r)' \\ &= (F \circ r)(b) - (F \circ r)(a) \\ &= F(z_2) - F(z_1). \end{aligned}$

In particular, if $\gamma$ is closed, then $z_1 = r(a) = r(b) = z_2$ , so that

$\displaystyle \begin{aligned} \oint_{\gamma} f(z)\, \mathrm dz &= F(z_2) - F(z_1) = 0. \end{aligned}$

Now if this condition holds, then let $\gamma_1$ and $\gamma_2$ be two distinct contours from $z_1$ to $z_2$ . Then $\gamma_1 + (-\gamma_2)$ is a contour from $z_1$ to itself, i.e. it is a closed contour, so that

$\displaystyle \begin{aligned} \oint_{\gamma_1} f(z)\, \mathrm dz - \oint_{\gamma_2} f(z)\, \mathrm dz &= \oint_{\gamma_1} f(z)\, \mathrm dz + \oint_{-\gamma_2} f(z)\, \mathrm dz \\ &= \oint_{\gamma_1 + (-\gamma_2)} f(z)\, \mathrm dz = 0,\end{aligned}$

implying that

$\displaystyle \oint_{\gamma_1} f(z)\, \mathrm dz = \oint_{\gamma_2} f(z)\, \mathrm dz.$

Finally, suppose the path-independence of the contour integral. Fix $z_0 \in D$ . For any $\zeta \in D$ , let $\gamma_{\zeta}$ be a contour from $z_0$ to $\zeta$ . Define

$\displaystyle F(\zeta) := \oint_{\gamma_{\zeta}} f(z)\, \mathrm dz.$

We claim that $F'(z_0) = f(z_0)$ . Fix $\epsilon > 0$ . By construction $F(z_0) = 0$ . Consider the straight-line contour $[z_0, z_0 +w]$ from $z_0$ to $z_0 +w$ . Then

$\begin{aligned} F(z_0 + w) - F(z_0) - f(z_0) \cdot w &= \oint_{[z_0, z_0+w]} f(z)\, \mathrm dz - 0 - \oint_{[z_0, z_0 + w]} f(z_0)\, \mathrm dz \\ &= \oint_{[z_0, z_0+w]} (f(z) - f(z_0))\, \mathrm dz. \end{aligned}$

Since $f$ is continuous at $z_0$ , for any $k > 0$ , there exists $\delta > 0$ such that

$|z| < \delta \quad \Rightarrow \quad |f(z) - f(z_0)| < k \cdot \epsilon.$

Define

$\displaystyle M := \max_{z \in [z_0, z_0+w]} |f(z_0) - f(z_0)| \leq k \cdot \epsilon,\quad L := |w|.$

Then for $0 < |w| < \delta$ , by the ML-inequality,

$\begin{aligned} |F(z_0 + w) - F(z_0) - f(z_0) \cdot w| &= \left| \oint_{[z_0, z_0+w]} (f(z) - f(z_0))\, \mathrm dz \right| \leq (k \cdot \epsilon) \cdot |w|.\end{aligned}$

Setting $k = 1/2$ yields the desired result.

Next time, we revisit multivariable integration in order to establish Green’s theorem, which we will need to prove the Cauchy-Goursat theorem, which states that sufficiently nice functions over sufficiently nice contours automatically have zero integrals.

—Joel Kindiak, 14 Aug 25, 2252H
October 31, 2025