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Wyatt's Notes — University

Singularities and Residue Theory

8.1 Isolated Singularities

Let z0z_0 be an isolated singularity of ff (i.e., ff is analytic in a punctured neighbourhood of z0z_0).

Classification by Laurent series:

  1. Removable singularity: an=0a_n = 0 for all n<0n \lt 0. Can be removed by redefining f(z0)=a0f(z_0) = a_0.
  2. Pole of order mm: a−m≠0a_{-m} \neq 0 and an=0a_n = 0 for n<−mn \lt -m. The principal part is finite.
  3. Essential singularity: infinitely many non-zero ana_n with n<0n \lt 0.

Proposition 8.1 (Riemann”s Removable Singularity Theorem). If ff is bounded near z0z_0Then z0z_0 is a removable singularity.

Proposition 8.2. z0z_0 is a pole of order mm if and only if lim⁡z→z0(z−z0)mf(z)\lim_{z \to z_0} (z - z_0)^m f(z) Exists and is non-zero.

Theorem 8.3 (Casorati-Weierstrass). If z0z_0 is an essential singularity of ffThen ff takes Values arbitrarily close to any complex number in every neighbourhood of z0z_0.

8.2 Classification with Worked Examples

Solution

Problem. Classify the singularities of f(z)=sin⁡zzf(z) = \frac{\sin z}{z}.

z=0z = 0: sin⁡z=z−z3/6+⋯\sin z = z - z^3/6 + \cdotsSo f(z)=1−z2/6+⋯f(z) = 1 - z^2/6 + \cdots. No negative powers, so z=0z = 0 is a removable singularity. f(0)=1f(0) = 1 by continuity.

Problem. Classify the singularities of f(z)=ez−1z2f(z) = \frac{e^z - 1}{z^2}.

z=0z = 0: ez−1=z+z2/2+⋯e^z - 1 = z + z^2/2 + \cdotsSo f(z)=1z+12+⋯f(z) = \frac{1}{z} + \frac{1}{2} + \cdots. Principal part is 1/z1/zSo z=0z = 0 is a simple pole with residue 11.

Problem. Classify the singularity of f(z)=e1/zf(z) = e^{1/z} at z=0z = 0.

e1/z=∑n=0∞1n! zn=1+1z+12z2+⋯e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n!\, z^n} = 1 + \frac{1}{z} + \frac{1}{2z^2} + \cdots

Infinitely many negative powers ⇒\Rightarrow z=0z = 0 is an essential singularity.

Problem. Classify the singularities of f(z)=z+1z3(z2+1)f(z) = \frac{z + 1}{z^3(z^2 + 1)}.

z=0z = 0: pole of order 33. z=iz = i: simple pole. z=−iz = -i: simple pole.

Problem. Determine the type of singularity of f(z)=zsin⁡zf(z) = \frac{z}{\sin z} at z=0z = 0.

sin⁡z=z−z3/6+⋯\sin z = z - z^3/6 + \cdotsSo f(z)=11−z2/6+⋯=1+z26+⋯f(z) = \frac{1}{1 - z^2/6 + \cdots} = 1 + \frac{z^2}{6} + \cdots.

No negative powers, so z=0z = 0 is a removable singularity with f(0)=1f(0) = 1.

8.3 Definition of the Residue

Definition. The residue of ff at an isolated singularity z0z_0 is the coefficient a−1a_{-1} In the Laurent expansion:

Res(f,z0)=a−1=12πi∫γf(z) dz\mathrm{Res}(f, z_0) = a_{-1} = \frac{1}{2\pi i}\int_\gamma f(z)\, dz

Where γ\gamma is a small positively oriented circle around z0z_0.

8.4 Computing Residues

For a simple pole at z0z_0:

Res(f,z0)=lim⁡z→z0(z−z0)f(z)\mathrm{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)

If f=g/hf = g/h where g(z0)≠0g(z_0) \neq 0, h(z0)=0h(z_0) = 0, h′(z0)≠0h'(z_0) \neq 0:

Res(f,z0)=g(z0)h′(z0)\mathrm{Res}(f, z_0) = \frac{g(z_0)}{h'(z_0)}

For a pole of order mm at z0z_0:

Res(f,z0)=1(m−1)!lim⁡z→z0dm−1dzm−1[(z−z0)mf(z)]\mathrm{Res}(f, z_0) = \frac{1}{(m-1)!}\lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}}\left[(z - z_0)^m f(z)\right]

Solution

Problem. Find the residue of f(z)=zz2+4z+3f(z) = \frac{z}{z^2 + 4z + 3} at each pole.

z2+4z+3=(z+1)(z+3)z^2 + 4z + 3 = (z + 1)(z + 3)So simple poles at z=−1z = -1 and z=−3z = -3.

At z=−1z = -1: Res=lim⁡z→−1zz+3=−12\mathrm{Res} = \lim_{z \to -1} \frac{z}{z + 3} = \frac{-1}{2}. At z=−3z = -3: Res=lim⁡z→−3zz+1=−3−2=32\mathrm{Res} = \lim_{z \to -3} \frac{z}{z + 1} = \frac{-3}{-2} = \frac{3}{2}.

Problem. Find the residue of f(z)=ez(z−1)2(z−2)f(z) = \frac{e^z}{(z - 1)^2(z - 2)} at each pole.

At z=1z = 1 (pole of order 22): Res=ddz[ezz−2]z=1=ez(z−2)−ez(z−2)2∣z=1=−e−e1=−2e\mathrm{Res} = \frac{d}{dz}\left[\frac{e^z}{z - 2}\right]_{z=1} = \frac{e^z(z - 2) - e^z}{(z-2)^2}\Big|_{z=1} = \frac{-e - e}{1} = -2e.

At z=2z = 2 (simple pole): Res=e2(2−1)2=e2\mathrm{Res} = \frac{e^2}{(2-1)^2} = e^2.

8.5 The Residue Theorem

Theorem 8.4 (Residue Theorem). If ff is analytic inside and on a simple closed positively Oriented contour γ\gamma except for isolated singularities z1,z2,…,znz_1, z_2, \ldots, z_n inside γ\gamma Then

∫γf(z) dz=2πi∑k=1nRes(f,zk)\int_\gamma f(z)\, dz = 2\pi i \sum_{k=1}^{n} \mathrm{Res}(f, z_k)

Proof. For each singularity zkz_kDraw a small circle γk\gamma_k around it. By Cauchy’s theorem Applied to the multiply connected region between γ\gamma and the γk\gamma_k:

∫γf dz=∑k=1n∫γkf dz=∑k=1n2πi⋅Res(f,zk)\int_\gamma f\, dz = \sum_{k=1}^n \int_{\gamma_k} f\, dz = \sum_{k=1}^n 2\pi i \cdot \mathrm{Res}(f, z_k). ■\blacksquare

8.6 Worked Examples: Residue Theorem

Solution

Problem 1. Evaluate ∫γezz(z−1)2 dz\int_\gamma \frac{e^z}{z(z-1)^2}\, dz where γ\gamma is ∣z∣=2|z| = 2.

Solution. Singularities inside γ\gamma: z=0z = 0 (simple pole) and z=1z = 1 (pole of order 22).

At z=0z = 0: Res=lim⁡z→0ez(z−1)2=1(−1)2=1\mathrm{Res} = \lim_{z \to 0} \frac{e^z}{(z-1)^2} = \frac{1}{(-1)^2} = 1.

At z=1z = 1: Res(f,1)=ddz[(z−1)2⋅ezz(z−1)2]z=1=ddz[ezz]z=1=ez⋅z−ezz2∣z=1=e−e1=0\mathrm{Res}(f, 1) = \frac{d}{dz}\left[(z-1)^2 \cdot \frac{e^z}{z(z-1)^2}\right]_{z=1} = \frac{d}{dz}\left[\frac{e^z}{z}\right]_{z=1} = \frac{e^z \cdot z - e^z}{z^2}\Big|_{z=1} = \frac{e - e}{1} = 0.

∫γf dz=2πi(1+0)=2πi\int_\gamma f\, dz = 2\pi i(1 + 0) = 2\pi i. ■\blacksquare

Problem 2. Evaluate ∫γ1z4+1 dz\int_\gamma \frac{1}{z^4 + 1}\, dz where γ\gamma is ∣z∣=2|z| = 2.

Solution. The poles are the fourth roots of −1-1: zk=eiπ/4+ikπ/2z_k = e^{i\pi/4 + ik\pi/2} for k=0,1,2,3k = 0, 1, 2, 3. All four lie inside ∣z∣=2|z| = 2.

Each is a simple pole with Res(f,zk)=14zk3\mathrm{Res}(f, z_k) = \frac{1}{4z_k^3}. Since zk4=−1z_k^4 = -1: zk−3=−zkz_k^{-3} = -z_kSo the sum equals −14∑zk=0-\frac{1}{4}\sum z_k = 0.

∫γdzz4+1=2πi⋅0=0\int_\gamma \frac{dz}{z^4 + 1} = 2\pi i \cdot 0 = 0. ■\blacksquare