Calculus of complex-valued functions

We briefly summarize the calculus of complex-valued functions with a real variable, which is needed to deal with differential equations and Fourier series. In the following, let $g=g(x)$ be a complex-valued function with a real variable $x\in \mathbb{R}.$ That is, $g:\mathbb{R}\to \mathbb{C}$. In short, we can apply all the calculus of real-valued functions to complex-valued functions by treating the imaginary unit $i=\sqrt{-1}$ as just another constant that happens to satisfy $i^2=-1$.

Continuity

$g(x)$ is said to be continuous at $x=x_0$ if

$$\underset{x\to x_0}{lim}g(x)=g(x_0)$$

which is equivalent to

$$\underset{x\to x_0}{lim}|g(x)-g(x_0)|=0,$$

or, to

$$\begin{array}{}\text{(eq:Ccont)}& \underset{x\to x_0}{lim}\mathrm{\Re }g(x)=\mathrm{\Re }g(x_0),\underset{x\to x_0}{lim}\mathrm{\Im }g(x)=\mathrm{\Im }g(x_0)\end{array}$$

where $\mathrm{\Re }g(x)$ and $\mathrm{\Im }g(x)$ are the real and imaginary parts of $g(x)$, respectively. According to (eq:Ccont), $g(x)$ is continuous at $x=x_0$ if and only if the real part $\mathrm{\Re }g(x)$ and the imaginary part $\mathrm{\Im }g(x)$ (which are real-valued functions) are continuous at $x=x_0$. Let $\overline{z}$ be the complex conjugate of $z\in \mathbb{C}$. Then $g$ is continuous if and only if $\overline{g}$ (the complex conjugate of $g(x)$) is continuous.

Differentiation

$g=g(x)$ is differentiable if and only if both $\mathrm{\Re }g$ and $\mathrm{\Im }g$ are differentiable. In this case, we have

$$\frac{d}{dx}g=\frac{d}{dx}\mathrm{\Re }g+i\frac{d}{dx}\mathrm{\Im }g.$$

Example. Let $\lambda \in \mathbb{R}$ be a constant. We have

$$e^{i\lambda x}=\cos (\lambda x)+i\sin (\lambda x)$$

and

$$\frac{d}{dx}{e}^{i\lambda x}=i\lambda e^{i\lambda x}=-\lambda \sin (\lambda x)+i\lambda \cos (\lambda x).$$

□

Note that the Mean Value Theorem does not hold for complex-valued functions!

Example. Consider $g(x)=e^{ix}$ on $I=[0,2\pi ]$. This is a continuous function. We have $g(0)=g(2\pi )=1$. We have $g^{\prime }(x)=i{e}^{ix}$, and hence $|g^{\prime }(x)|=1$. Thus, there exists no $\xi \in I$ such that $g^{\prime }(\xi )=0$. Thus, Rolle's Theorem (a special case of the Mean Value Theorem) does not hold for this function. □

Integration

Consider the definite integral ${\int }_a^{b}g(x)dx$ of $g=g(x)$ on $a\le x\le b$. It can be defined by

$${\int }_a^{b}g(x)dx={\int }_a^b\mathrm{\Re }g(x)dx+i{\int }_a^b\mathrm{\Im }g(x)dx.$$

Or, equivalently, we can define by the limit of the Riemann sum over the partition $\mathrm{\Delta }:a=x_0<x_1<\cdots <x_m=b$,

$${\sigma }_{\mathrm{\Delta }}=\sum _{k=0}^{m-1}g({\xi }_k)(x_{k+1}-x_k)$$

as the mesh of $\mathrm{\Delta }$ approaches 0. Here ${\xi }_k$ is an arbitrary point such that $x_k\le {\xi }_k\le x_{k+1}$.

Many of the properties of the definite integral of real-valued functions also hold as long as they make sense for complex-valued functions.

1. If $g(x)$ is continuous on $[a,b]$, then ${\int }_a^{b}g(x)dx$ exists, and $$|{\int }_a^{b}g(x)dx|\le {\int }_a^b|g(x)|dx.$$
2. If $g(x)$ is continuous on $[a,b]$, then the function $G(x)$ defined by $$G(x)={\int }_a^{x}g(t)dt$$ is a primitive function (anti-derivative) of $g(x)$. That is, $$\frac{d}{dx}G(x)=g(x).$$ Conversely, if $G_1(x)$ is a primitive function of $g(x)$, then $${\int }_a^{b}g(x)dx={[G_1(x)]}_a^b=G_1(b)-G_1(a).$$

Example. Let $\gamma$ be a non-zero complex number.

$${\int }_a^b{e}^{\gamma x}dx={\left[\frac{e^{\gamma x}}{\gamma }\right]}_a^b=\frac{e^{\gamma b}-e^{\gamma a}}{\gamma }.$$

□

Integration by parts, by substitution, Arzelà's theorem, etc., also hold for complex-valued functions. Improper integrals are also defined.