Complex function theory survival guide

Not every student may have had enough complex analysis to appreciate the manipulations we must carry out when -- for example -- calculating Green's functions in this course; others may have forgotten most of what they've learnt about this subject. Hence, I list the most important results of complex function theory without proof.

An analytic function is a complex function which can be differentiated an infinite number of times. It turns out that a complex function which is differentiable, satisfies the Cauchy-Riemann equations. In order to formulate these equations, we first introduce some notation. A point in the complex plane is given as $$z = x + {\rm i}y.$$ A complex function $f(z)$ may then be written as $$f(z) = u(x,y) + {\rm i}v(x,y).$$ $u$ is the real, and $v$ the imaginary part of the complex function. The Cauchy-Riemann equations are then $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} ; \phantom{xxx} \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$$ It turns out that this condition is sufficiently strong to ensure infinite differentiability: a complex function which can be differentiated once, can be differentiated an infinite number of times. Functions that satisfy this requirement are called analytic.

We often deal with integrations over closed curves ('contours'). It can be shown that the integral of an analytic function taken over such a contour gives zero: $$\oint_\Gamma f(z) dz = 0,$$ where $\Gamma$ denotes the contour. We adopt the convention that in complex integration, the contour is always traversed in the anti-clockwise direction. Reversing the direction reverses the sign of the result (which in the case of an analytic function has no effect, as the result of the integration is 0).

We often deal with functions having singularities. Point-like singularities are called poles. We say that a function $f$ has a pole of order $n$ in the point $a$ on the complex plane if $(z-a)^n f(z)$ is analytic in $a$, but $(z-a)^{n-1} f(z)$ is not. Now suppose we expand $f$ around $a$ as a series expansion in $z-a$, including negative powers: $$f(z) = \sum_{n=-\infty}^\infty (z-a)^n f_n(a) .$$ The residue of $f$ in $a$, denoted as ${\rm res}_{a} f$ is defined as the coefficient of $(z-a)^{-1}$ in this expansion. For a pole of order one, also a called a simple pole, the residue of $f$ is given as $\lim_{z\rightarrow a} (z-a)f(a)$. In general, for an isolated pole, the residue is defined as $${\text{res}}_a f(z) = \frac{1}{(n-1)!} \lim_{z\rightarrow a} \frac{d^{n-1}}{dz^{n-1}} \left[ (z-a)^n f(z)\right].$$

The most important result of complex analysis that we shall be using frequently, is about functions with a set of isolated poles $a_1, a_2, \ldots, a_k$ within the closed contour $\Gamma$. We then have $$\oint_\Gamma f(z) dz = 2 \pi {\rm i}\sum_{j=1}^k {\rm res}_{a_j} f.$$ This is the so-called residue theorem.

Let us consider an example. We calculate the integral $$\oint \frac{1}{1+z^2} dz$$ over a circle of radius 2 around the origin. The contour contains the points $\pm {\rm i}$. We note that $$\frac{1}{1+z^2} =\frac{{\rm i}}{2} \frac{1}{z+{\rm i}} - \frac{{\rm i}}{2} \frac{1}{z-{\rm i}} .$$ Both terms yield a standard integral with a simple pole, and working them out using the residue theorem yields the value $$\oint \frac{1}{1+z^2} dz = \frac{{\rm i}}{2} (2\pi {\rm i}) - \frac{{\rm i}}{2} (2\pi {\rm i}) = 0.$$

Another result is important for the cases we will be dealing with. Consider a semi-circle with radius $R$ in the upper complex plane, and centred around 0. We call this circle $\Gamma^+$. Then, Jordan's lemma says that if the function $f(z)$ is bounded on $\Gamma^+$, the integral $$\int_{\Gamma^+} e^{{\rm i}kz} f(z) dz$$ is finite, and for $R\rightarrow \infty$ it approaches zero.

As an example, we calculate $$\int_{-\infty}^\infty \frac{e^{{\rm i}kx}}{x - a} dx,$$ where $a$ is a complex number with a positive imaginary part and $k$ is real. We now evaluate the integral over the contour $\Gamma$ shown in figure A.1. Because of Jordan's Lemma, this integral is equal to the integral over the real axis only. Using the residue theorem, we immediately have $$\int_{-\infty}^\infty \frac{e^{{\rm i}kx}}{x - a} dx = \oint \frac{e^{{\rm i}kx}}{x - a} dx - \int_{\Gamma^+} \frac{e^{{\rm i}kx}}{x - a} dx.$$ Now we use Jordan's lemma which says that the second term vanishes, and the integral over the closed contour can be evaluated using the residue theorem: $$\oint \frac{e^{{\rm i}kx}}{x - a} dx = 2\pi {\rm i}e^{ika}.$$

FIGURE A.1: The contour for evaluating an integral along the real axis.

Finally, we consider the integral $$\int \frac{f(x)}{x} dx$$ over the real axis. This integral is not well defined at $x=0$ unless $f$ vanishes there (and is continuously differentiable at $x=0$). Now let us consider the same integral but running just above the real axis: $$\int \frac{f(x+{\rm i}\epsilon)}{x+{\rm i}\epsilon} dx,$$ where $\epsilon$ is small. Supposing again that $f$ is regular, the small $\epsilon$ does not change its value significantly, but we have avoided the singularity at $x=0$. For $x$ not very close to zero, the integral is approximately equal to the one running over the real axis, and we should focus on what happens near $x=0$. Let us work out the imaginary part of $1/(x+{\rm i}\epsilon)$: $${\rm Im} \left( \frac{1}{x+{\rm i}\epsilon}\right) = \frac{1}{2{\rm i}} \left( \frac{1}{x+{\rm i}\epsilon} - \frac{1}{x-{\rm i}\epsilon} \right) = -\frac{\epsilon}{x^2 + \epsilon^2}.$$ The right hand side of this function is a narrow peak centred around $x=0$. Its integral is $-\pi$. Therefore, for small $\epsilon$, the integral will be $-\pi \delta (x)$. All in all, we see that we can write $$\frac{1}{x+{\rm i}\epsilon} = 1/x {\text{ (for }} x {\text{ away from zero) }} - {\rm i}\pi \delta(x).$$ The first part of this expression is called the principal value and denoted as ${\mathcal P}$: $$\frac{1}{x+{\rm i}\epsilon} = {\mathcal P} \left(\frac{1}{x}\right) - {\rm i}\pi \delta(x).$$ More precisely $${\mathcal P} \left(\frac{1}{x}\right) = 1/x {\text{ for }} |x|>\epsilon, \epsilon \rightarrow 0.$$ For completeness, we write down the principal value integral over a function $f(x)$ having a singularity in $a$ is: $${\mathcal P} \int f(x) dx = \lim_{\epsilon\downarrow 0} \int_{-\infty}^{a-\epsilon} f(x) dx + \int_{a+\epsilon}^\infty f(x) dx.$$