Residues and Fourier Transforms

Cauchy's theorem says that the integral of a holomorphic function around a loop is zero. When the function has isolated singularities inside the loop, the integral is not zero, but it is still easy: each singularity contributes $2\pi i$ times a single number, its residue. This turns contour integration into a calculating machine. Real integrals that resist every trick of first-year calculus fall in a few lines, and so do the Fourier transforms of 4A.5 The Fourier Transform.

The same idea, applied to $f'/f$, counts zeros: the integral of $f'/f$ around a loop is $2\pi i$ times the number of zeros inside, minus the number of poles. This is the argument principle, and it is the first theorem in the guide where an integral computes a whole number with topological meaning. Engineers use it every day, in the form of the Nyquist stability criterion.

By the end of this chapter you will be able to:

• classify isolated singularities and compute residues;
• state and prove the residue theorem, and use it to evaluate real integrals;
• compute Fourier transforms by shifting contours, and explain why analyticity of $f$ makes $\hat f$ decay exponentially;
• state and prove the argument principle and Rouché's theorem;
• explain the Nyquist criterion as an application of the argument principle, and use it on an example.

3.1Will the loop oscillate?

3.2Zeros, poles and Laurent series

Let $f$ be holomorphic on a punctured disc $0 < |z - a| < r$. The point $a$ is an isolated singularity, and there are three possibilities.

• Removable: $f$ extends holomorphically to $a$. By Riemann's theorem this happens as soon as $f$ is bounded near $a$ (Exercise 3.12).
• Pole: $|f(z)| \to \infty$ as $z \to a$. Then $f(z) = (z - a)^{-m}g(z)$ for some integer $m \geq 1$, the order of the pole, and $g$ holomorphic with $g(a) \neq 0$. A function holomorphic except for poles is meromorphic.
• Essential: neither. Near an essential singularity $f$ comes arbitrarily close to every complex value (the Casorati–Weierstrass theorem); $e^{1/z}$ at $0$ is the standard example.

In all three cases $f$ has a Laurent expansion

converging on the punctured disc. It comes from Cauchy's formula applied on an annulus, with the inner circle contributing the negative powers. The singularity is removable if no negative powers occur, a pole of order $m$ if the most negative is $(z - a)^{-m}$, and essential if there are infinitely many.

The residue is the only coefficient that survives integration around $a$, because $\oint(z - a)^n\,dz$ is $2\pi i$ for $n = -1$ and $0$ for every other integer (5A.2 Cauchy’s Theorem and Its Consequences). At poles it is computed without finding the whole series:

• at a simple pole, $\operatorname{res}_af = \lim_{z\to a}(z - a)f(z)$; in particular, if $f = g/h$ with $g(a) \neq 0$ and $h$ having a simple zero at $a$, then $\operatorname{res}_af = \frac{g(a)}{h'(a)}$;
• at a pole of order $m$, $\operatorname{res}_af = \frac{1}{(m-1)!}\lim_{z\to a}\frac{d^{m-1}}{dz^{m-1}}\big[(z - a)^mf(z)\big]$.

For example, $\frac{1}{1 + z^2}$ has simple poles at $\pm i$ with residues $\frac{1}{2i}$ and $-\frac{1}{2i}$, and $\frac{e^z}{z^3}$ has residue $\frac12$ at $0$, the coefficient of $z^2$ in $e^z$.

3.3The residue theorem

3.3.1Real integrals

The method for an integral over the real line: close the segment $[-R, R]$ with a semicircle in the upper half-plane, apply the residue theorem, and show that the semicircle contributes nothing as $R \to \infty$ (Figure 3.1).

Integrals over a period, $\int_0^{2\pi}R(\cos\theta, \sin\theta)\,d\theta$ with $R$ rational, go the other way: substitute $z = e^{i\theta}$, so $\cos\theta = \frac12(z + z^{-1})$, $\sin\theta = \frac{1}{2i}(z - z^{-1})$ and $d\theta = \frac{dz}{iz}$, and the integral becomes a contour integral around the unit circle (Exercise 3.14).

3.4Fourier transforms by moving contours

We use the convention of 4A.5 The Fourier Transform: $\hat f(\xi) = \int_{-\infty}^\infty f(x)e^{-2\pi ix\xi}\,dx$.

These examples show a general principle. In 4A.5 The Fourier Transform smoothness of $f$ was traded for polynomial decay of $\hat f$: each derivative buys a power of $|\xi|$. Analyticity buys much more.

The function $\frac{1}{1 + x^2}$ is holomorphic in the strip $|\operatorname{Im}z| < 1$, with poles on its edges at $\pm i$, and its transform $\pi e^{-2\pi|\xi|}$ decays at exactly the borderline rate $b = 1$. The Gaussian is entire, and its transform decays faster than every exponential. The Paley–Wiener theorem (Stein–Shakarchi, chapter 4) completes the picture: $\hat f$ vanishes outside $[-M, M]$ exactly when $f$ extends to an entire function of exponential type $2\pi M$.

3.5The argument principle

In words: as $z$ goes once around $\gamma$, the argument of $f(z)$ increases by $2\pi$ times the number of zeros minus poles inside. The answer is an integer, so it doesn't change under small perturbations, which is the source of its power.

For example, $z^5 + 3z + 1$ has all five of its zeros in $|z| < 2$, since $|z^5| = 32 > 7 \geq |3z + 1|$ on $|z| = 2$; and exactly one in $|z| < 1$, since there $|3z| = 3 > 2 \geq |z^5 + 1|$ (Exercise 3.16).

3.5.1The Nyquist criterion

3.6History

Cauchy introduced residues in 1826 and developed the calculus of residues over the following years; Pierre Alphonse Laurent's expansions date from 1843 (Weierstrass had found them earlier but not published). Eugène Rouché's theorem appeared in 1862. The decay of Fourier transforms of analytic functions was made precise by Raymond Paley and Norbert Wiener in their 1934 book Fourier Transforms in the Complex Domain. Harry Nyquist, at Bell Telephone Laboratories, published "Regeneration Theory" in the Bell System Technical Journal in January 1932, to analyse the stability of feedback amplifiers; it founded the frequency-domain approach to the stability of control systems.

3.7Exercises

Solution

(a) Pole of order $2$: $\frac{d}{dz}(z + i)^{-2}\big|_{z=i} = -2(2i)^{-3} = -\frac{2}{-8i} = \frac{1}{4i} = -\frac i4$. (b) $\frac{1}{\cos0} = 1$. (c) $\frac{z}{4z^3} = \frac{1}{4z^2}$ at $z = e^{i\pi/4}$: $\frac{1}{4i} = -\frac i4$. (d) $e^{1/z} = \sum\frac{1}{n!z^n}$, residue $1$ (an essential singularity).

Solution

The poles in the upper half-plane are $e^{i\pi/4}$ and $e^{3i\pi/4}$, with residues $\frac{1}{4z^3} = \frac{z}{4z^4} = -\frac z4$, i.e. $-\frac14e^{i\pi/4}$ and $-\frac14e^{3i\pi/4}$. Their sum is $-\frac14\cdot i\sqrt2$, and $2\pi i$ times it is $\frac{\pi\sqrt2}{2} = \frac{\pi}{\sqrt2}$. The semicircle contributes at most $\frac{\pi R}{R^4 - 1} \to 0$.

Solution

With $z = e^{i\theta}$ the integral is $\oint_{|z|=1}\frac{dz}{iz\,(a + \frac12(z + z^{-1}))} = \frac2i\oint\frac{dz}{z^2 + 2az + 1}$. The roots are $-a \pm\sqrt{a^2 - 1}$; only $z_+ = -a + \sqrt{a^2 - 1}$ is inside the circle (their product is $1$). The residue there is $\frac{1}{z_+ - z_-} = \frac{1}{2\sqrt{a^2 - 1}}$, so the integral is $\frac2i\cdot2\pi i\cdot\frac{1}{2\sqrt{a^2 - 1}} = \frac{2\pi}{\sqrt{a^2 - 1}}$.

Solution

$(s + 1)^3 = -K$ gives $s + 1 = K^{1/3}e^{i\pi(2k+1)/3}$, $k = 0, 1, 2$: angles $60°$, $180°$, $300°$. The root at $180°$ is $s = -1 - K^{1/3}$, always stable. The others have real part $-1 + \frac12K^{1/3}$, positive exactly when $K > 8$. For $K = 12$, $K^{1/3} \approx 2.289$, so $s \approx 0.145 \pm 1.983i$: an oscillation growing slowly, at a frequency close to $\sqrt3 \approx 1.73$, the crossover frequency of the Nyquist plot.

Solution

$w^2 = V^2\big(1 - \frac{a^2}{z^2}\big)^2 + \frac{iV\Gamma}{\pi z}\big(1 - \frac{a^2}{z^2}\big) - \frac{\Gamma^2}{4\pi^2z^2}$. The only $\frac1z$ term is $\frac{iV\Gamma}{\pi z}$, so $\oint w^2\,dz = 2\pi i\cdot\frac{iV\Gamma}{\pi} = -2V\Gamma$, and $X - iY = \frac{i\rho}{2}(-2V\Gamma) = -i\rho V\Gamma$. So $X = 0$ and $Y = \rho V\Gamma$. That there is no drag at all is d'Alembert's paradox: in an ideal fluid without viscosity, steady flow exerts no drag on any body.

Solution

$e^{-4\pi^2t\xi^2} = g(c\xi)$ with $g(\xi) = e^{-\pi\xi^2}$ and $c = 2\sqrt{\pi t}$. If $\hat h = g$ then $\hat h(c\,\cdot)$ is the transform of $\frac1ch(x/c)$, so the inverse transform is $\frac{1}{2\sqrt{\pi t}}e^{-\pi x^2/4\pi t} = \frac{1}{\sqrt{4\pi t}}e^{-x^2/4t}$. Its transform decays like $e^{-4\pi^2t\xi^2}$, faster than $e^{-2\pi b|\xi|}$ for every $b$, so, as in the ahead box after Theorem 3.6, the inversion integral converges for every complex $x$ and defines an entire function.