A Taylor series lives on a disk, and stops at the first singular point. Around a singular point, or in a ring between two of them, we need negative powers too. Expanding with the geometric series in the right variable is almost always the tool, as in the previous lesson.
Key idea
If $f$ is analytic on an annulus $r < |z - a| < R$, then on that annulus
$$f(z) = \sum_{n = -\infty}^{\infty} c_n(z - a)^n, \qquad c_n = \frac{1}{2\pi i}\oint_C\frac{f(w)}{(w - a)^{n + 1}}\,dw,$$
for any circle $C$ in the annulus. The series is unique for that annulus, but a different annulus (around the same center) gets a different series. The basic rule: $\frac{1}{z - c} = -\frac{1}{c}\sum_{n \ge 0}\left(\frac{z}{c}\right)^n$ where $|z| < |c|$, and $\frac{1}{z - c} = \frac{1}{z}\sum_{n \ge 0}\left(\frac{c}{z}\right)^n$ where $|z| > |c|$.
Why it works
For the two expansions of $\frac{1}{z - c}$, factor out whichever of $z$ and $c$ is bigger, so the ratio inside the geometric series has modulus less than 1: $\frac{1}{z - c} = \frac{-1}{c}\cdot\frac{1}{1 - z/c}$ or $\frac{1}{z}\cdot\frac{1}{1 - c/z}$. For a general $f$, Cauchy's formula on the boundary of the annulus splits $f$ into a piece analytic inside the outer circle (a Taylor series in $z - a$) and a piece analytic outside the inner circle (a series in $\frac{1}{z - a}$).
Worked examples
Worked example: Two rings, two series
Expand $f(z) = \dfrac{1}{z(z - 1)}$ on $0 < |z| < 1$ and on $|z| > 1$.
- Partial fractions: $f = \frac{1}{z - 1} - \frac{1}{z}$.
- On $0 < |z| < 1$: $\frac{1}{z - 1} = -\sum_{n \ge 0}z^n$, so $f = -\frac{1}{z} - 1 - z - z^2 - \cdots$.
- On $|z| > 1$: $\frac{1}{z - 1} = \sum_{n \ge 0}\frac{1}{z^{n + 1}}$, so $f = \frac{1}{z^2} + \frac{1}{z^3} + \frac{1}{z^4} + \cdots$ (the $\frac{1}{z}$ terms cancel).
Answer: $-\frac{1}{z} - 1 - z - \cdots$ and $\frac{1}{z^2} + \frac{1}{z^3} + \cdots$
Worked example: A middle ring
Find the coefficient of $z^{-1}$ and of $z^{0}$ for $f(z) = \dfrac{1}{(z - 1)(z - 2)}$ on $1 < |z| < 2$.
- $f = \frac{1}{z - 2} - \frac{1}{z - 1}$.
- $|z| > 1$: $-\frac{1}{z - 1} = -\frac{1}{z} - \frac{1}{z^2} - \cdots$. $|z| < 2$: $\frac{1}{z - 2} = -\frac{1}{2} - \frac{z}{4} - \cdots$.
- Coefficient of $z^{-1}$: $-1$. Coefficient of $z^0$: $-\frac{1}{2}$.
Answer: $-1$ and $-\frac{1}{2}$
Worked example: An essential singularity
Find the Laurent series of $z^2e^{1/z}$ on $|z| > 0$ and its coefficient of $z^{-1}$.
- $e^{1/z} = \sum_{n \ge 0}\frac{1}{n!\,z^n}$, valid for all $z \ne 0$.
- $z^2e^{1/z} = z^2 + z + \frac{1}{2} + \frac{1}{6z} + \frac{1}{24z^2} + \cdots$.
- The coefficient of $z^{-1}$ is $\frac{1}{6}$; infinitely many negative powers appear.
Answer: $\frac{1}{6}$
- Split the function into partial fractions $\frac{A}{z - c}$.
- For each pole $c$, compare $|c|$ with the ring: if the ring lies inside $|z| < |c|$, expand in powers of $\frac{z}{c}$; if outside, in powers of $\frac{c}{z}$.
- Add the expansions and collect the coefficient you need.
Common mistake
Using the wrong geometric series for the ring. On $|z| > 1$, writing $\frac{1}{z - 1} = -\sum z^n$ is wrong: that series diverges there. The ratio must be small: on $|z| > 1$ use powers of $\frac{1}{z}$. A quick check: plug in a point of the ring (such as $z = 2$ for $|z| > 1$) and ask whether the terms of your series shrink.