# 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 $\begin{split} \int_{\gamma} \alpha f(z) + \beta g(z) \, dz & = \alpha \int_0^1 f(\gamma(t)) \gamma'(t) \, dt \\ & \phantom{=} + \beta \int_0^1 g(\gamma(t)) \gamma'(t) \, dt \end{split}$ $$\blacksquare$$

### 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 $$\gamma(t) = \gamma_k ((t - t_k)/(t_{k+1} - t_k))$$ 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 $\begin{split} \ell(\gamma) &= \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} \left| \frac {\gamma_k'\left( \frac{t-t_k}{t_{k+1} - t_k} \right)} {t_{k+1} - t_k} \, dt \right| \end{split}$ 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 [$$\dagger$$].

‌ 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 ($$f'$$ being continuous) 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 if 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: $\left[a \to a + \frac{h}{3} \to a + \left(1+e^{-i\pi/3} \right)\frac{h}{3}\to a + 2\frac{h}{3} \to a+h\right]$

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.