Computing Residues Using Laurent Series: A Beginner’s Guide

Introduction

Have you ever wondered how mathematicians work with functions that "blow up" at certain points? In complex analysis, we use Laurent series to express functions with singularities (points where they become infinite or undefined). From these series, we can extract something called the residue, which plays a crucial role in evaluating contour integrals.

In this post, we’ll learn how to compute residues using Laurent series by working through two examples:

  1. f(z)=e1/zf(z) = e^{1/z}, a function with an essential singularity at z=0z = 0.
  2. g(z)=ln⁡(1+z)1+z2g(z) = \frac{\ln(1+z)}{1+z^2}, a function with isolated singularities that we will analyze.

No prior knowledge of complex analysis is needed—just a curiosity for math!

What is a Laurent Series?

A Laurent series is like a Taylor series but allows for negative powers of zz. It takes the form:

f(z)=∑n=−∞∞cn(z−a)n.f(z) = \sum_{n=-\infty}^{\infty} c_n (z-a)^n.

where:

  • cnc_n are coefficients,
  • aa is the center of expansion,
  • Negative powers of (z−a)(z-a) represent singular behavior.

Residue Definition: The residue of f(z)f(z) at z=az = a is the coefficient c−1c_{-1} in the Laurent series expansion of f(z)f(z) around z=az = a.

The residue is key to the Residue Theorem, which allows us to evaluate complex integrals using simple calculations.

Example 1: Finding the Residue of f(z)=e1/zf(z) = e^{1/z} at z=0z = 0

We begin with the function:

f(z)=e1/z.f(z) = e^{1/z}.

Step 1: Expand the Function

We use the Maclaurin series for exe^x:

ex=∑n=0∞xnn!.e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.

Substituting x=1/zx = 1/z, we get:

e1/z=∑n=0∞(1/z)nn!=∑n=0∞1n!z−n.e^{1/z} = \sum_{n=0}^{\infty} \frac{(1/z)^n}{n!} = \sum_{n=0}^{\infty} \frac{1}{n!} z^{-n}.

Step 2: Identify the Residue

From the Laurent series:

e1/z=10!z0+11!z−1+12!z−2+13!z−3+…e^{1/z} = \frac{1}{0!} z^0 + \frac{1}{1!} z^{-1} + \frac{1}{2!} z^{-2} + \frac{1}{3!} z^{-3} + \dots

The coefficient of z−1z^{-1} is 11, so:

Res(f,0)=1.\text{Res}(f, 0) = 1.

This tells us that f(z)f(z) has a residue of 1 at z=0z = 0.

Example 2: Laurent Series for g(z)=ln⁡(1+z)1+z2g(z) = \frac{\ln(1+z)}{1+z^2}

Now, let's consider:

g(z)=ln⁡(1+z)1+z2.g(z) = \frac{\ln(1+z)}{1+z^2}.

This function has singularities where 1+z2=01+z^2 = 0, which happens at:

z=±i.z = \pm i.

We will expand the function in a Laurent series around z=0z = 0.

Step 1: Expand ln⁡(1+z)\ln(1+z)

We use the Maclaurin series for ln⁡(1+z)\ln(1+z):

ln⁡(1+z)=∑n=1∞(−1)n+1znn,for ∣z∣<1.\ln(1+z) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{z^n}{n}, \quad \text{for } |z| < 1.

So:

ln⁡(1+z)=z−z22+z33−z44+…\ln(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \dots

Step 2: Divide by 1+z21+z^2

Using the geometric series expansion:

11+z2=∑n=0∞(−1)nz2n,\frac{1}{1+z^2} = \sum_{n=0}^{\infty} (-1)^n z^{2n},

we multiply:

g(z)=(z−z22+z33−z44+… )⋅∑n=0∞(−1)nz2n.g(z) = (z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \dots) \cdot \sum_{n=0}^{\infty} (-1)^n z^{2n}.

Step 3: Identify the Residue

To find the residue, we look for the coefficient of z−1z^{-1}. Notice that all terms contain only nonnegative powers of zz, meaning there is no z−1z^{-1} term in the Laurent series.

Thus, the residue of g(z)g(z) at z=0z = 0 is:

Res(g,0)=0.\text{Res}(g, 0) = 0.

This tells us that g(z)g(z) has no contribution to contour integrals at z=0z = 0.

P.S. Laurent Series at z=iz = i for g(z)g(z)

Now, let’s expand g(z)g(z) around z=iz = i. To do this, we substitute w=z−iw = z - i (so that z=w+iz = w + i).

Since:

1+z2=(z−i)(z+i),1 + z^2 = (z-i)(z+i),

we rewrite:

g(z)=ln⁡(1+z)(z−i)(z+i).g(z) = \frac{\ln(1+z)}{(z-i)(z+i)}.

Expanding ln⁡(1+z)\ln(1+z) at z=iz = i, we write:

ln⁡(1+z)=ln⁡(1+i)+∑n=1∞(−1)n+1(z−i)nn.\ln(1+z) = \ln(1+i) + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(z-i)^n}{n}.

Using partial fraction decomposition:

1(z−i)(z+i)=Az−i+Bz+i,\frac{1}{(z-i)(z+i)} = \frac{A}{z-i} + \frac{B}{z+i},

solving for AA, we get:

A=12i.A = \frac{1}{2i}.

Thus, the residue at z=iz = i is:

Res(g,i)=ln⁡(1+i)2i.\text{Res}(g, i) = \frac{\ln(1+i)}{2i}.

Final Thoughts

Residues are incredibly useful in complex analysis and help evaluate integrals in physics and engineering. If you want to practice, try finding the Laurent series and residues for ezz\frac{e^z}{z} at z=0z = 0!

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