# Introduction

The main goal of this chapter is to derive the fundamental theorem of calculus for functions of a complex variable. This theorem characterizes the relation between functions and their primitives with the help of integrals. A version of this theorem for functions of a real variable is the following:

### Theorem – Fundamental Theorem of Calculus (Real Analysis).

‌ Let $$I$$ be an open interval of $$\mathbb{R},$$ $$f:I \to \mathbb{R}$$ be a continuous function and $$a \in I.$$ A function $$g: I \to \mathbb{R}$$ is a primitive of $$f$$ if and only if it satisfies $\forall \, x \in I, \; g(x) = g(a) + \int_a^x f(t) \, dt.$

### Proof.

‌ Suppose that the function $$g$$ satisfies the integral equation of the theorem. For any $$x \in I$$ and any real number $$h$$ such that $$x+h \in I,$$ $\begin{split} \frac{g(x+h) - g(x)}{h} &= \frac{1}{h}\int_{x}^{x+h} f(t) \, dt \\ &= \frac{1}{h}\int_x^{x+h} f(x) \, dt + \frac{1}{h}\int_x^{x+h} (f(t) - f(x)) \, dt \\ &= \, f(x) + \frac{1}{h}\int_x^{x+h} (f(t) - f(x)) \, dt, \\ \end{split}$ Let $$\epsilon > 0;$$ by continuity of $$f$$ at $$x,$$ there is a $$\delta > 0$$ such that $\forall \, t \in I, \; (|t-x| \leq \delta \, \Rightarrow \, |f(t) - f(x)| < \epsilon)$ thus if $$|h| < \delta,$$ $\left| \frac{g(x+h) - g(x)}{h} - f(x) \right| \leq \frac{1}{|h|} |h| \times \epsilon = \epsilon.$ The difference quotient tends to $$f(x)$$ when $$h$$ tends to zero: $$g'(x)$$ exists and is equal to $$f(x).$$

Conversely, suppose that $$e: I \to \mathbb{R}$$ is a primitive of $$f.$$ The difference $$d$$ between $$e$$ and the function $g: x \in I \mapsto e(a) + \int_a^x f(t) \, dt$ is zero at $$a$$ and has a zero derivative on $$I.$$ By the mean value theorem, for any $$x \in I$$ such that $$x\neq a,$$ there is a $$b \in I$$ such that $\frac{d(x) - d(a)}{x - a} = d'(b) = 0,$ hence $$d(x) = d(a) = 0$$ and therefore $$e = g.$$ $$\blacksquare$$

# Paths

### Definition – Path.

‌ A path $$\gamma$$ is a continuous function from $$[0,1]$$ to $$\mathbb{C}.$$ If $$A$$ is a subset of the complex plane, $$\gamma$$ is a path of $$A$$ if additionally $$\gamma([0,1]) \subset A.$$

### Definition – Image of a Path.

‌ The image or trajectory or trace of the path $$\gamma$$ is the image $$\gamma([0,1])$$ of the interval $$[0,1]$$ by the function $$\gamma.$$

### Definition – Path Endpoints.

‌ The complex numbers $$\gamma(0)$$ and $$\gamma(1)$$ are the initial point and terminal point of $$\gamma$$ – they are its endpoints; the path $$\gamma$$ joins its initial and terminal points. The path is closed if the initial and terminal point are the same. The paths $$\gamma_1, \dots, \gamma_n$$ are consecutive if for $$k = 1 ,\dots, n-1,$$ the terminal point of $$\gamma_k$$ is the initial point of $$\gamma_{k+1}.$$

### Example – Oriented Line Segment.

‌ The oriented line segment (or simply oriented segment) with initial point $$a \in \mathbb{C}$$ and terminal point $$b \in \mathbb{C}$$ is denoted $$[a \to b]$$ and defined as $[a \to b]: t \in [0,1] \mapsto (1-t) a + t b.$ Its image is the line segment $$[a,b].$$ Representation of the oriented line segment $$[0 \to 2+i]$$

### Example – Oriented Circle.

‌ The oriented circle of radius one centered at the origin traversed once in the positive sense (counterclockwise) is denoted $$[\circlearrowleft]$$ and defined as $[\circlearrowleft]: t \in [0,1] \to e^{i2\pi t}.$ The circle of radius $$r \geq 0$$ centered at $$c \in \mathbb{C}$$ traversed $$n\in \mathbb{Z}^*$$ times in the positive sense is the path $c + r[\circlearrowleft]^n: t \in [0,1] \to c + r e^{i2\pi n t}.$ Its image is the circle centered on $$c$$ with radius $$r;$$ its initial and terminal points are both $$c+r,$$ hence it is closed. Representation of the oriented circle $$[\circlearrowleft]$$

### Definition – Open (Path-)Connected Sets.

‌ An open subset $$\Omega$$ of the complex-plane is (path-)connected if for any points $$x$$ and $$y$$ of $$\Omega,$$ there is a path of $$\Omega$$ that joins $$x$$ and $$y.$$

### Definition – Reverse Path.

‌ The reverse (or opposite) of the path $$\gamma$$ is the path $$\gamma^{\leftarrow}$$ defined by $\forall \, t \in [0,1], \; \gamma^{\leftarrow}(t) = \gamma(1-t).$

### Definition – Path Concatenation.

‌ Let $$t_0 = 0 < t_1 < \dots < t_{n-1} < t_{n} = 1$$ be a partition of the interval $$[0,1].$$ The concatenation of consecutive paths $$\gamma_1, \dots, \gamma_n$$ associated to this partition is the path $$\gamma$$ denoted $\gamma_1 \, |_{t_1} \, \cdots \, |_{t_{n-1}} \, \gamma_{n}$ such that $\forall\, k \in \{1, \dots, n\}, \; \gamma|_{[t_{k-1},t_{k}]} = \gamma_{k}\left( \frac{t-t_{k-1}}{t_{k} - t_{k - 1}}\right).$ If the partition of $$[0,1]$$ is uniform, that is, if $\forall \, k \in \{0,\dots, n\}, \; t_{k} = k/n,$ we denote the concatenated path with the simpler notation $\gamma_1 \, | \, \cdots \, | \, \gamma_n.$

### Example – Oriented Polyline.

‌ An oriented polyline (or piecewise linear path) is the concatenation of consecutive oriented line segments. When the associated partition of $$[0,1]$$ is uniform, we use the notation $[a_0 \to a_1 \to \dots \to a_{n-1} \to a_n] = [a_0 \to a_1] \, | \, \cdots \, | \, [a_{n-1} \to a_n].$

### Definition – Rectifiable Path.

‌ A path $$\gamma:[0,1] \to \mathbb{C}$$ is rectifiable if the function $$\gamma$$ is piecewise continuously differentiable.
Given the definition of piecewise continuously differentiable, the following alternate characterization is plain:

### Theorem – Continuously Differentiable Decomposition.

‌ A path $$\gamma:[0,1] \to \mathbb{C}$$ is rectifiable if and only if there are consecutive continuously differentiable paths $$\gamma_1, \dots, \gamma_n$$ and a partition $$(t_0, \dots, t_n)$$ of the interval $$[0,1]$$ such that $\gamma = \gamma_1 \, |_{t_1} \, \cdots \, |_{t_{n-1}} \, \gamma_n.$
We characterized initially connected sets via merely continuous paths. However, when such sets are open, we can use rectifiable paths instead:

### Lemma – Connectedness & Rectifiable Paths.

‌ An open subset $$\Omega$$ of the complex plane is connected if and only if every pair of points of $$\Omega$$ may be joined by a rectifiable path of $$\Omega.$$

### Proof.

‌ If any pair of points of $$\Omega$$ can be joined by a rectifiable path of $$\Omega,$$ then $$\Omega$$ is connected. Conversely, assume that a (merely continuous) path $$\gamma$$ of $$\Omega$$ joins $$x$$ and $$y.$$ Its image $$\gamma([0,1])$$ is a compact subset of $$\Omega$$ – as the image of a compact set by a continuous function – thus the distance $$r$$ between $$\gamma([0,1])$$ and the closed set $$\mathbb{C} \setminus \Omega$$ is positive. Additionally, the function $$\gamma$$ is uniformly continuous – as a continuous function with a compact domain of definition; there is a positive integer $$n$$ such that $\forall \, t \in [0,1], \; \forall \, s \in [0,1], \; (|t - s| \leq 1/n \, \Rightarrow \, |\gamma(t) - \gamma(s)| < r).$ For any $$k \in \{0,\dots,n\},$$ the point $$\gamma(k/n)$$ belongs to $$\Omega;$$ the path $$\mu$$ defined as $\mu = [\gamma(0) \to \dots \to \gamma(k/n) \to \dots \to \gamma(1)]$ is rectifiable and joins $$x$$ and $$y.$$ Now, for any $$t \in [0,1],$$ let $$k \in \{0,\dots, n-1\}$$ be such that $$t \in [k/n, (k+1)/n].$$ We have $|\mu(t) - \gamma(k/n)| \leq |\gamma((k+1)/n) - \gamma(k/n)| < r,$ therefore $$\mu$$ is a path of $$\Omega.$$ $$\blacksquare$$

# Line Integrals

### Definition – Length of a Rectifiable Path.

‌ The length of a rectifiable path $$\gamma$$ is the nonnegative real number $\ell(\gamma) = \int_0^1 |\gamma'(t)| \, dt.$

### Example – Length of an Oriented Segment.

‌ The oriented segment $$[a \to b]$$ is continuously differentiable and thus rectifiable. For any $$t\in[0,1],$$ $$[a\to b]'(t) = b - a,$$ hence its length is $\ell([a \to b]) = \int_0^1 |b - a| \, dt = |b - a|.$

### Example – Length of an Oriented Circle.

‌ The oriented circle $$c+ r[\circlearrowleft]^n$$ centered at $$c$$ with radius $$r\geq 0$$ traversed $$n$$ times in the positive sense is continuously differentiable and thus rectifiable. For any $$t \in [0,1],$$ $[c+ r[\circlearrowleft]^n]'(t) = (i2\pi n) r e^{i2\pi n t},$ hence the length of this path is $\ell(c+ r[\circlearrowleft]^n) = \int_0^1 |(i2\pi n) r e^{i2\pi n t}| \, dt = \int_0^1 |2\pi n r| \, dt = 2\pi r \times |n|.$ It differs from the length of its circle image – which is $$2\pi r$$ – unless the circle is traversed exactly once in the positive or negative sense.

### Definition – Line Integral.

‌ The line integral along a rectifiable path $$\gamma$$ of a complex-valued function $$f$$ defined and continuous on the image of $$\gamma$$ is the complex number defined by $\int_{\gamma} f(z) \, dz = \int_0^1 f(\gamma(t)) \gamma'(t) \, dt.$

### Remark – Undefined Integrands.

‌ In the definitions of the length of $$\gamma$$ and of the integral along $$\gamma$$, the integrands $|\gamma'(t)| \; \mbox{ and } \; f(\gamma(t))\gamma'(t)$ may be undefined for some values of $$t$$ if $$\gamma$$ is merely rectifiable. However it’s not an issue since they are always defined almost everywhere (and integrable).

### Remark – Integral Notation.

‌ It’s sometimes handy to use the notation $\int_{\gamma} f(z) \, |dz| = \int_0^1 f(\gamma(t)) |\gamma'(t)| dt.$ which is similar to the one used for line integrals. With this convention, we have for example $\ell(\gamma) = \int_{\gamma} |dz|.$

### Example – Integration along an Oriented Segment.

‌ The line integral of the continuous function $$f: [a,b] \mapsto \mathbb{C}$$ along the oriented segment $$[a \to b]$$ is $\begin{split} \int_{[a \to b]} f(z) \, dz = &\, \int_{0}^1 f((1-t) a + t b) (b-a)\, dt \\ = &\, (b-a) \int_{0}^1 f((1-t) a + t b) \, dt. \end{split}$

### Example – Integration along an Oriented Circle.

‌ The line integral of a continuous function $$f: \{z \in \mathbb{C} \; | \; |z|=1\} \to \mathbb{C}$$ on the oriented circle $$[\circlearrowleft]$$ is $\begin{split} \int_{[\circlearrowleft]} f(z) \, dz = &\, \int_{0}^1 f(e^{i2\pi t }) (i2\pi e^{i2\pi t} dt) \\ = &\, i \int_{0}^1 f(e^{i2\pi t }) e^{i2\pi t}\, (2\pi dt) \\ = &\, i \int_{0}^{2\pi} f(e^{i\theta})e^{i\theta} d\theta. \end{split}$

### Theorem – Complex-Linearity of the Line Integral.

‌ Let $$\gamma$$ be a rectifiable path. For any $$\alpha, \beta \in \mathbb{C}$$ and any continuous functions $$f$$ and $$g$$ defined on the image of $$\gamma$$, $\int_{\gamma} \alpha f(z) + \beta g(z) \, dz = \alpha \int_{\gamma} f(z) \, dz + \beta \int_{\gamma} g(z) \, dz.$

### Proof.

‌ Since by definition of the line integral $\int_{\gamma} \alpha f(z) + \beta g(z) \, dz = \int_0^1 (\alpha f(\gamma(t)) + \beta g(\gamma(t))) \gamma'(t) dt,$ the complex-linearity of the integral on $$[0,1]$$ provides $\begin{split} \int_{\gamma} \alpha f(z) + \beta g(z) \, dz & = \alpha \int_0^1 f(\gamma(t)) \gamma'(t) \, dt \\ & \phantom{=} + \beta \sum_k \int_0^1 g(\gamma(t)) \gamma'(t) \, dt \end{split}$ $$\blacksquare$$

### Theorem – Integration along a Reverse Path.

‌ For any rectifiable path $$\gamma$$, $\ell(\gamma^{\leftarrow}) = \ell(\gamma).$ For any continuous function $$f: A \subset \mathbb{C} \to \mathbb{C}$$ defined on the image of $$\gamma$$, $\int_{\gamma^{\leftarrow}} f(z) \, dz = - \int_{\gamma} f(z) \, dz.$

### Proof.

‌Since $$\gamma^{\leftarrow}(t) = \gamma(1-t)$$, the length of the opposite of $$\gamma$$ satisfies $\ell(\gamma^{\leftarrow}(t)) = \int_0^1 |(\gamma^{\leftarrow})'(t)| \, dt = \int_0^1 |-\gamma'(1-t)| \, dt.$ The change of variable $$t \mapsto 1- t$$ yields $\ell(\gamma^{\leftarrow}(t)) = \int_0^1 |\gamma'(t)| \, dt = \ell(\gamma).$ Similarly, $\begin{split} \int_{\gamma^{\leftarrow}} f(z) \, dz &= \int_0^1 f(\gamma^{\leftarrow}(t)) (\gamma^{\leftarrow})'(t) \, dt \\ &= \int_0^1 f(\gamma(1-t)) (-\gamma'(1-t))\, dt \\ &= \int_0^1 f(\gamma(t)) (-\gamma'(t))\, dt \\ &= - \int_{\gamma} f(z) \, dz \end{split}$ $$\blacksquare$$

### Theorem – Integration along Concatenation of Paths.

‌ Let $$A$$ be a subset of $$\mathbb{C}$$. Let $$\gamma_1, \dots, \gamma_n$$ be consecutive rectifiable paths of $$A$$ and let $$\gamma$$ be their concatenation $\gamma = \gamma_1 \, |_{t_1} \, \cdots \, |_{t_{n-1}} \, \gamma_{n}.$ The length of $$\gamma$$ satisfies $\ell(\gamma) = \sum_{k=1}^n \ell(\gamma_k).$ For any continuous function $$f: A \subset \mathbb{C} \to \mathbb{C}$$ defined on the image of $$\gamma$$, $\int_{\gamma} f(z) \, dz = \sum_{k=1}^n \int_{\gamma_k} f(z) \, dz.$

### Proof.

‌ Since by definition $$\gamma(t) = \gamma_k ({(t - t_k)}/{(t_k - t_{k-1})})$$ whenever $$t \in [t_k, t_{k+1}]$$, the decomposition $\begin{split} \ell(\gamma) &= \int_0^1 |\gamma'(t)| \, dt \\ &= \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} |\gamma'(t)| \, dt \end{split}$ provides $\begin{split} \ell(\gamma) &= \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} \left| \frac {\gamma_k'\left( \frac{t-t_k}{t_{k+1} - t_k} \right)} {t_k - t_{k-1}} \, dt \right| \end{split}$ and the changes of variables $$t \in [t_k, t_{k+1}] \mapsto \frac{t-t_k}{t_{k+1} - t_k}$$ yield $\begin{split} \ell(\gamma) &= \sum_{k=1}^n \int_0^1 |\gamma_k'(t)| \, dt \\ &= \sum_{k=1}^n \ell(\gamma_k) \end{split}$

Similarly, $\begin{split} \int_{\gamma} f(z) dz &= \int_0^1 f(\gamma(t)) \gamma'(t) \, dt \\ &= \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} f(\gamma(t)) \gamma'(t) \, dt \\ &= \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} f\left(\gamma_k\left( \frac{t-t_k}{t_{k+1} - t_k} \right) \right) \frac{\gamma_k'\left( \frac{t-t_k}{t_{k+1} - t_k} \right)}{t_k - t_{k-1}} \, dt \\ &= \sum_{k=1}^n \int_0^1 f(\gamma_k(t)) \gamma_k'(t) \, dt \\ &= \sum_{k=1}^n \int_{\gamma_k} f(z) \, dz \end{split}$ $$\blacksquare$$

### Theorem – M-L Inequality.

‌ For any rectifiable path $$\gamma$$ and any continuous function $$f: A \subset \mathbb{C} \to \mathbb{C}$$ defined on the image of $$\gamma$$, $\left| \int_{\gamma} f(z) \, dz \right| \leq \left( \max_{z \in \gamma([0,1])} |f(z)| \right) \times \ell(\gamma).$

### Proof.

By definition of the line integral $\begin{split} \left| \int_{\gamma} f(z) \, dz \right| =&\, \left| \int_0^1 f(\gamma(t)) \gamma'(t) \, dt \right| \\ \leq &\, \int_0^1 |f(\gamma(t))| |\gamma'(t)| \, dt \\ \leq &\, \left( \max_{t \in [0,1]} |f(\gamma(t))| \right) \times \int_0^1 |\gamma'(t)| \, dt \\ = &\, \left( \max_{z \in \gamma([0,1])} |f(z)| \right) \times \ell(\gamma). \end{split}$ $$\blacksquare$$

A practical consequence of the M-L inequality:

### Corollary – Convergence in Line Integrals.

‌ For any rectifiable path $$\gamma$$ and any sequence of continuous function $$f_n: A \subset \mathbb{C} \to \mathbb{C}$$ defined on the image of $$\gamma$$ which converges uniformly to the function $$f,$$ $\lim_{n \to +\infty} \int_{\gamma} f_n(z) \, dz = \int_{\gamma} f(z) \, dz.$

### Proof.

‌ The linearity of the line integral and the M-L inequality provide $\begin{split} \left| \int_{\gamma} f_n(z) \, dz - \int_{\gamma} f(z) \right| & = \left| \int_{\gamma} (f_n(z) - f(z)) \, dz \right| \\ & \leq \left( \max_{z \in \gamma([0,1])} |f_n - f(z)| \right) \times \ell(\gamma) \end{split}$ which yields the desired result. $$\blacksquare$$

### Theorem – Invariance By Reparametrization.

‌ Let $$\gamma:[0,1] \to \mathbb{C}$$ be a continuously differentiable path. Let $$\phi:[0,1] \to [0,1]$$ be an increasing $$C^1$$-diffeomorphism – a continuously differentiable function such that $$\phi(0) = 0,$$ $$\phi(1) = 1$$ and $$\phi'(t)>0$$ for any $$t \in [0,1].$$ The following statements hold:
• The path $$\mu = \gamma \circ \phi$$ is a continuously differentiable path.

• It has the same initial point, terminal point and image as $$\gamma.$$

• The length of $$\mu$$ and $$\gamma$$ are identical.

• For any continuous function $$f: \gamma([0,1]) \to \mathbb{C},$$ $\int_{\mu} f(z) \, dz = \int_{\gamma} f(z) \, dz.$

### Proof.

‌ The function $$\mu$$ is continuously differentiable as the composition of continuously differentiable functions. We have $\mu(0) = \gamma(\phi(0)) = \gamma(0), \; \mu(1) = \gamma(\phi(1)) = \gamma(1),$ hence the endpoints of $$\gamma$$ and $$\mu$$ are identical. The function $$\phi$$ is a bijection from $$[0,1]$$ into itself, therefore $\mu([0,1]) = \gamma (\phi([0,1])) = \gamma ([0,1])$ and the images of $$\gamma$$ and $$\mu$$ are identical.

The length of $$\mu$$ is $\ell(\mu) = \int_0^1 |\mu'(t)|\, dt = \int_{0}^1 |\gamma' (\phi(t)) \phi'(t)| \, dt = \int_{0}^1 |\gamma' (\phi(t))| \, \phi'(t) dt$ The change of variable $$s = \phi(t)$$ provides $\int_{0}^1 |\gamma' (\phi(t))| \, \phi'(t) dt = \int_0^1 |\gamma'(s)| \, ds,$ hence the lengths of $$\gamma$$ and $$\mu$$ are equal. We also have $\int_{\mu} f(z) \, dz = \int_{0}^1 (f \circ \mu)(t) \mu'(t)\,dt = \int_{0}^1 (f \circ \gamma)(\phi(t)) \gamma'(\phi(t))\,(\phi'(t) dt).$ The same change of variable leads to $\int_{\mu} f(z) \, dz = \int_{0}^1 (f \circ \gamma)(s) \gamma'(s)\,ds = \int_{\gamma} f(z) \, dz,$ which concludes the proof. $$\blacksquare$$

### Definition – Image of a Path by a Function.

‌ Let $$\gamma:[0,1] \to \mathbb{C}$$ be a path and $$f:A \subset \mathbb{C} \to \mathbb{C}$$ be a continuous function defined on the image of $$\gamma$$. The image of $$\gamma$$ by $$f$$ is the path $$f \circ \gamma.$$

### Theorem – Change of Variable in Line Integrals.

‌ Let $$\Omega$$ be an open subset of $$\mathbb{C},$$ let $$\gamma$$ be a rectifiable path of $$\Omega$$ and let $$f:\Omega \to \mathbb{C}$$ be a holomorphic function. The path $$f \circ \gamma$$ is rectifiable and for any continuous function $$g: A \subset \mathbb{C} \to \mathbb{C}$$ defined on the image of $$f \circ \gamma$$, $\int_{f \circ \gamma} g(z) \, dz = \int_{\gamma} g(f(w)) f'(w) \, dw.$

### Proof.

‌Let $$\gamma_1 \, |_{t_1} \, \dots \, |_{t_{n-1}} \, \gamma_n$$ be a continously differentiable decomposition of $$\gamma.$$ We have $f \circ \gamma = f \circ \gamma_1 \, |_{t_1} \, \dots \, |_{t_{n-1}} \, f \circ \gamma_n,$ and for any $$k \in \{1,\dots,n\},$$ by the chain rule, the function $$f \circ \gamma_{k}$$ is continuously differentiable hence the path $$f \circ \gamma$$ is rectifiable.

Moreover, $\begin{split} \int_{\gamma} g(f(w)) f'(w) \, dw = &\, \int_0^1 g(f(\gamma(t)) f'(\gamma(t)) \gamma'(t) \, dt \\ = &\, \int_0^1 g(f(\gamma(t)) (f \circ \gamma)'(t) \, dt \\ = &\, \int_{f \circ \gamma} g(w) \, dw \end{split}$ $$\blacksquare$$

# Primitives

### Definition – Primitive.

‌ Let $$f: \Omega \to \mathbb{C}$$ where $$\Omega$$ is an open subset of $$\mathbb{C}.$$ A primitive (or antiderivative) of $$f$$ is a holomorphic function $$g: \Omega \to \mathbb{C}$$ such that $$g' = f.$$

### Theorem – Fundamental Theorem of Calculus (Complex Analysis).

‌ Let $$\Omega$$ be an open connected subset of $$\mathbb{C},$$ $$f: \Omega \to \mathbb{C}$$ be a continuous function and let $$a \in \Omega.$$ A function $$g:\Omega \to \mathbb{C}$$ is a primitive of $$f$$ if and only if for any $$z \in \Omega$$ and any rectifiable path $$\gamma$$ of $$\Omega$$ that joins $$a$$ and $$z,$$ $g(z) = g(a) + \int_{\gamma} f(w) \, dw.$

### Proof.

‌Let $$g$$ be a primitive of $$f$$ and $$\gamma$$ be a rectifiable path of $$\Omega$$ that joins $$a$$ and $$z.$$ Let $$\gamma = \gamma_1 \, |_{t_1} \, \dots \, |_{t_{n-1}} \, \gamma_n$$ be a continuously differentiable decomposition of $$\gamma$$. For any $$k \in \{1,\dots, n\},$$ the function $\phi: t \in [0,1] \mapsto g(\gamma_{k}(t))$ is differentiable as a composition of real-differentiable functions, with $\phi'(t) = dg_{\gamma_{k}(t)} (\gamma'_{k}(t)) = g'(\gamma_{k}(t)) \gamma_{k}'(t).$ The function $$\phi'$$ is continuous hence by the fundamental theorem of calculus (from real analysis) applied to the real and imaginary parts of $$\phi'$$ on $$\left]0,1\right[,$$ we have for any positive number $$\epsilon$$ smaller than $$1,$$ $\phi(1-\epsilon) - \phi(\epsilon) = \int_{\epsilon}^{1-\epsilon} \phi'(t) \, dt,$ and thus by continuity of $$\phi$$ and $$\phi'$$ $\phi(1) - \phi(0) = \int_0^1 \phi'(t) \, dt,$ which is equivalent to $g(\gamma_{k}(1)) - g(\gamma_{k}(0)) = \int_0^1 g'(\gamma_{k}(t)) \gamma_{k}'(t) \, dt = \int_{\gamma_{k}} f(w) \, dw.$ The sum of these equations for all $$k \in \{1,\dots, n\}$$ provides $g(z) - g(a) = \int_{\gamma} f(w) \, dw.$ Conversely, assume that $$g$$ satisfies the theorem property. Let $$\gamma$$ be a rectifiable path of $$\Omega$$ that joins $$a$$ and $$z$$ and let $$r>0$$ be such that the open disk centered at $$z$$ with radius $$r$$ is included in $$\Omega.$$ Consider the concatenation $$\mu$$ of $$\gamma$$ and of the oriented segment $$[z \to z+h]$$ for $$h$$ such that $$|h| < r.$$ It is a rectifiable path of $$\Omega,$$ hence $\begin{split} g(z+h) = &\, g(a) + \int_{\mu} f(w) \, dw \\ = &\, g(a) + \int_{\gamma} f(w)\, dw + h \int_0^1 f(z + t h) \, dt \\ = &\, g(z) + h \int_0^1 f(z + t h) \, dt \end{split}$ hence $\frac{g(z+h) - g(z)}{h} = \int_0^1 f(z + t h) \, dt.$ The right-hand side of this equation converges to $$f(z)$$ by continuity when $$h$$ goes to zero, therefore $$g$$ is a primitive of $$f.$$ $$\blacksquare$$

### Corollary – Existence of Primitives [$$\dagger$$].

‌ Let $$\Omega$$ be an open connected subset of $$\mathbb{C}.$$ The function $$f: \Omega \to \mathbb{C}$$ has a primitive if and only if it is continuous and for any closed rectifiable path $$\gamma$$ $\int_{\gamma} f(z) \, dz = 0.$

### Proof – Existence of Primitives.

‌ If the function $$f$$ has primitives, it is the derivative of a holomorphic function, thus it is continuous. Additionally, for any closed rectifiable path $$\gamma$$ of $$\Omega,$$ the fundamental theorem of calculus provides $g(\gamma(1)) = g(\gamma(0)) + \int_{\gamma} f(w)\, dw,$ hence as $$\gamma(1) = \gamma(0),$$ $\int_{\gamma} f(w) \, dw = 0.$ Conversely, assume that any such integral is zero. Select any $$a$$ in $$\Omega$$ and define for any point $$z$$ in $$\Omega$$ and any rectifiable path $$\gamma$$ of $$\Omega$$ that joins them the function $g(z) = g(a) + \int_{\gamma} f(w) \, dw.$ This definition is non-ambiguous: if we select a different path $$\mu,$$ the difference between the right-hand sides of the definitions would be $\left(g(a) + \int_{\gamma} f(w) \, dw\right) - \left(g(a) + \int_{\mu} f(w) \, dw\right) = \int_{\gamma \, | \, \mu^{\leftarrow}} f(w) \, dw = 0$ as $$\gamma \, | \, \mu^{\leftarrow}$$ is a closed rectifiable path of $$\Omega.$$ Consequently, $$g$$ is uniquely defined and by the fundamental theorem of calculus, it is a primitive of $$f.$$ $$\blacksquare$$

### Corollary – Set of Primitives.

‌ Let $$\Omega$$ be an open connected subset of $$\mathbb{C}$$ and let $$f: \Omega \to \mathbb{C}.$$ If $$g:\Omega \to \mathbb{C}$$ is a primitive of $$f,$$ the function $$h: \Omega \to \mathbb{C}$$ is also a primitive of $$f$$ if and only iff it differs from $$g$$ by a constant.

### Proof.

‌ It is clear that a function $$h$$ that differs from $$g$$ by a constant is a primitive of $$f.$$ Conversely, if $$g$$ and $$h$$ are both primitives of $$f,$$ $$g - h$$ is a primitive of the zero function. The fundamental theorem of calculus shows that for any $$a$$ and $$z$$ in $$\Omega$$ and any rectifiable path $$\gamma$$ of $$\Omega$$ that joins them, $g(z) - h(z) = g(a) - h(a) + \int_{\gamma} 0 \, dw = g(a) - h(a)$ hence their difference is a constant. $$\blacksquare$$

### Corollary – Integration by Parts [$$\dagger$$].

‌ Let $$\Omega$$ be an open connected subset of $$\mathbb{C}$$ and let $$\gamma$$ be a rectifiable path of $$\Omega.$$ For any pair of holomorphic functions $$f: \Omega \to \mathbb{C}$$ and $$g: \Omega \to \mathbb{C},$$ $\int_{\gamma} f' g (z) \, dz = [fg(\gamma(1)) - fg(\gamma(0))] - \int_{\gamma} f g' (z) \, dz.$

### Proof.

The derivative of the function $$fg$$ is $$f'g + fg'.$$ It is continuous as a sum and product of continuous functions thus the fundamental theorem of calculus provides $fg(\gamma(1)) = fg(\gamma(0)) + \int_{\gamma} (f'g + fg')(z) \, dz,$ which is equivalent to the conclusion of the corollary. $$\blacksquare$$

### Remark & Definition – Variation of a Function on a Path.

‌ The difference between the value of a function $$f$$ at the terminal value and at the initial value of a path $$\gamma$$ may be denoted $$[f]_{\gamma}$$: $[f]_{\gamma} = f(\gamma(1)) - f(\gamma(0)).$ With this convention, the formula that relates a function $$f$$ and its primitive $$g$$ is $[g]_{\gamma} = \int_{\gamma} f(z) \, dz$ and the integration by parts formula becomes $\int_{\gamma} f' g (z) \, dz = [fg]_{\gamma} - \int_{\gamma} f g' (z) \, dz.$

# Appendix – A Better Theory of Rectifiability

## Rectifiable Paths

The definition we used so far for “rectifiable” is a conservative one. In this section, we come up with a more general definition of the concept that still meets the requirements for the definition of line integrals.

To “rectify” a path (from Latin rectus “straight” and facere “to make”) is to straighten – or by extension to compute its length, which is a trivial operation once a path has been straightened.

The general definition of the length of a path does not require line integrals. Instead, consider any partition $$(t_0,\dots,t_n)$$ of the interval $$[0,1]$$ and the path $$\mu= \mu_1 \, |_{t_1} \, \dots \,|_{t_{n-1}} \mu_n$$ where $\mu_{k}(t) = (1-t) \gamma(t_{k-1}) + t \gamma(t_{k}).$ We may define the length of such a combination of straight lines as $\ell(\mu) = \sum_{k=1}^{n-1} |\gamma(t_{k+1}) - \gamma(t_{k})|.$ As the straight line is the shortest path between two points, this number should provide a lower bound of the length of $$\gamma.$$ On the other hand, using finer partitions of the interval $$[0,1]$$ should also provide better approximations of the length of $$\gamma.$$ Following this idea, we may define the length of $$\gamma$$ as the supremum of the length of $$\mu$$ for all possible partitions of $$[0,1]$$: $\ell(\gamma) = \sup \left\{ \sum_{k=0}^{n-1} |\gamma(t_{k+1}) - \gamma(t_{k})| \; \left| \; \vphantom{\sum_k^k} n\in \mathbb{N}^*, \, t_0=0 < \dots < t_n =1 \right. \right\}$

Not every path has a finite length; those who have are by definition rectifiable. In general, a function $$\gamma:[0,1] \mapsto \mathbb{C}$$ whose length is finite – even if it is not continuous – is of bounded variation.

## The Line Integral

To define line integrals along the path $$\gamma,$$ it is enough that $$\gamma$$ is of bounded variation. For any such function $$\gamma,$$ we may build a (complex-valued, Borel) measure on $$[0,1]$$ denoted $$d\gamma.$$ This measure is defined by its integral of any continuous function $$\phi:[0,1]\to \mathbb{C},$$ as a limit of Riemann(-Stieltjes) sums $\int_{[0,1]} \phi \, d\gamma = \lim \sum_{m=0}^{n-1} \phi (t_m) (\gamma(t_{m+1}) - \gamma(t_m)).$ The limit is taken over the partitions of the interval $$[0,1]$$ with $\max \, \{|t_{m+1} - t_{m}| \; | \; m \in \{0,\dots, n-1\}\} \to 0.$ The line integral of a continuous function $$f: \gamma([0,1]) \to \mathbb{C}$$ is then defined by $\int_{\gamma} f(z) \, dz = \int_{[0,1]} (f \circ \gamma) \, d\gamma.$ The total variation $$|d\gamma|$$ of $$d\gamma$$ is the positive measure defined by $|d\gamma|(A) = \sup_{\mathfrak{P}} \sum_{B \in \mathfrak{P}} |d\gamma(B)|$ where the supremum is taken over all finite partitions $$\mathfrak{P}$$ of $$A$$ into measurable sets. This measure provides an integral expression for the length of $$\gamma$$: $\ell(\gamma) = \int_{[0,1]} |d\gamma|.$

## A Non-Rectifiable Curve

The Koch snowflake (Koch 1904) is an example of a continuous curve which is is nowhere differentiable; it is also a non-rectifiable closed path. It is defined as the limit of a sequence of polylines $$\gamma_n.$$ The first element of this sequence is an oriented equilateral triangle: $\gamma_1 = [0 \to 1 \to e^{i\pi3} \to 0].$ Then, $$\gamma_{n+1}$$ is defined as a transformation of $$\gamma_n$$: every oriented line segment $$[a \to a+h]$$ that composes $$\gamma_n$$ is replaced by the polyline: $\left[a \to a + \frac{h}{3} \to a + \left(1+e^{-i\pi/3} \right)\frac{h}{3}\to a + 2\frac{h}{3} \to a+h\right]$ Image of the Koch snowflake, first iteration. Image of the Koch snowflake, second iteration. Image of the Koch snowflake, third iteration. Image of the Koch snowflake, fourth iteration

The Koch snowflake $$\gamma$$ is defined as the limit of the $$\gamma_n$$ sequence. The geometric construction yields that for any $$n$$ greater than zero, $\forall \, t \in [0,1], \; |\gamma_{n+1}(t) - \gamma_n(t)| \leq \left(\frac{1}{3}\right)^{n} \frac{\sqrt{3}}{2}.$ As $$\sum_{p=0}^{+\infty} \left( \frac{1}{3}\right)^p = \frac{1}{1-1/3} = \frac{3}{2},$$ for any positive integer $$p$$ we have $\forall \, t \in [0,1], \; |\gamma_{n+p}(t) - \gamma_n(t)| \leq \left(\frac{1}{3}\right)^{n} \frac{3}{2}\frac{\sqrt{3}}{2}.$ The sequence $$\gamma_n$$ is a Cauchy sequence in the space of continuous and complex-valued functions defined on $$[0,1];$$ its uniform limit exists and is also continuous.

On the other hand, the curve is not rectifiable. First, the definition of the sequence $$\gamma_n$$ makes it plain that every iteration increases the initial length of the path by one-third: $\ell(\gamma_n) = 3 \times \left(\frac{4}{3}\right)^{n-1}.$ The length of $$\gamma_n$$ tends to $$+\infty$$ when $$n \to +\infty.$$ Now, every point at the junction of the segments of the polyline $$\gamma_n$$ also belongs to the Koch snowflake; more precisely $\forall \, m \in \{0,\dots, 3 \times4^{n-1}\}, \; \gamma\left(\frac{m}{3 \times 4^{n-1}}\right) = \gamma_n\left(\frac{m}{3 \times 4^{n-1}}\right).$ Therefore $\ell(\gamma) \geq \sum_{m=0}^{3 \times 4^{n-1} -1} \left| \gamma\left(\frac{m+1}{3 \times 4^{n-1}}\right) - \gamma\left(\frac{m}{3 \times 4^{n-1}}\right)\right| = \ell(\gamma_n)$ and thus $$\ell(\gamma) = +\infty$$: the path $$\gamma$$ is not rectifiable.