Calculus of complex-valued functions

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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

\[\lim_{x\to x_0}g(x) = g(x_0)\]

which is equivalent to

\[\lim_{x\to x_0}|g(x) - g(x_0)| = 0,\]

or, to

\[\lim_{x\to x_0}\Re{g(x)} = \Re{g(x_0)}, ~~~\lim_{x\to x_0}\Im{g(x)} = \Im{g(x_0)}\tag{eq:Ccont}\]

where \(\Re{g(x)}\) and \(\Im{g(x)}\) 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 \(\Re{g(x)}\) and the imaginary part \(\Im{g(x)}\) (which are real-valued functions) are continuous at \(x = x_0\). Let \(\bar{z}\) be the complex conjugate of \(z\in\mathbb{C}\). Then \(g\) is continuous if and only if \(\bar{g}\) (the complex conjugate of \(g(x)\)) is continuous.

Differentiation

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

\[\frac{d}{dx}g = \frac{d}{dx}\Re{g} + i\frac{d}{dx}\Im{g}.\]

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

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

and

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

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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 \(g'(x) = ie^{ix}\), and hence \(|g'(x)| = 1\). Thus, there exists no \(\xi\in I\) such that \(g'(\xi) = 0\). Thus, Rolle's Theorem (a special case of the Mean Value Theorem) does not hold for this function. □

Integration

Consider the definite integral \(\int_a^bg(x)\,dx\) of \(g = g(x)\) on \(a \leq x \leq b\). It can be defined by

\[\int_a^bg(x)\,dx = \int_a^b\Re{g(x)}\,dx + i \int_a^b\Im{g(x)}\,dx.\]

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

\[\sigma_{\Delta} = \sum_{k=0}^{m-1}g(\xi_k)(x_{k+1} - x_{k})\]

as the mesh of \(\Delta\) approaches 0. Here \(\xi_k\) is an arbitrary point such that \(x_k \leq \xi_k \leq x_{k+1}\).

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 \(\int_a^bg(x)\,dx\) exists, and \[\left|\int_a^bg(x)dx\right| \leq \int_a^b|g(x)|dx.\]
  2. If \(g(x)\) is continuous on \([a,b]\), then the function \(G(x)\) defined by \[G(x) = \int_a^xg(t)\,dt\] 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 \[\int_a^bg(x)dx = \left[G_1(x)\right]_a^b = G_1(b) - G_1(a).\]

Example. Let \(\gamma\) be a non-zero complex number.
\[\int_a^be^{\gamma x}dx = \left[\frac{e^{\gamma x}}{\gamma}\right]_a^b = \frac{e^{\gamma b} - e^{\gamma a}}{\gamma}.\]
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Integration by parts, by substitution, Arzelà's theorem, etc., also hold for complex-valued functions. Improper integrals are also defined.

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