Laurent Series & Singularities
Laurent Series & Singularities
When $f$ has an isolated singularity at $z_0$, it cannot be expanded in a Taylor series there, but it has a Laurent series $\sum_{n=-\infty}^\infty a_n(z-z_0)^n$. The nature of the singularity is determined by the principal part $\sum_{n<0}a_n(z-z_0)^n$: removable (finitely many terms with $n<0$, all zero), pole of order $m$ (finitely many terms, smallest $n=-m$), or essential (infinitely many). The residue $a_{-1}$ at a pole is crucial for contour integration.
Laurent Series & Singularity Classification
At isolated singularity $z_0$, $f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n$ for $0<|z-z_0|
Residue Theorem
If $f$ is analytic inside and on $\gamma$ except at isolated singularities $z_1,\ldots,z_k$ inside $\gamma$: $$\oint_\gamma f(z)\,dz=2\pi i\sum_{j=1}^k\text{Res}(f,z_j).$$ This vastly generalizes Cauchy's formula and is the key tool for evaluating real integrals by contour integration.
Example 1
Find the Laurent series of $\frac{1}{z(z-1)}$ for $0<|z|<1$.
Solution: Partial fractions: $\frac{1}{z(z-1)}=\frac{1}{z-1}-\frac{1}{z}=-\frac{1}{z}+\frac{1}{z-1}$. For $|z|<1$: $\frac{1}{z-1}=-\frac{1}{1-z}=-\sum_{n=0}^\infty z^n$. So $f(z)=-\frac{1}{z}-\sum_{n=0}^\infty z^n=\cdots-z^{-1}-1-z-z^2-\cdots$. Residue at $z=0$: $a_{-1}=-1$; residue at $z=1$: from $f(z)=\frac{1}{z}\cdot\frac{1}{z-1}$, $\text{Res}(f,1)=\lim_{z\to 1}(z-1)f(z)=\frac{1}{1}=1$.
Example 2
Evaluate $\oint_{|z|=2}\frac{z}{(z-1)^2(z+1)}\,dz$.
Solution: Singularities inside $|z|=2$: pole of order 2 at $z=1$; simple pole at $z=-1$. $\text{Res}(f,-1)=\lim_{z\to-1}(z+1)f(z)=\frac{-1}{(-2)^2}=-1/4$. $\text{Res}(f,1)=\lim_{z\to 1}\frac{d}{dz}\left[(z-1)^2\frac{z}{(z-1)^2(z+1)}\right]=\frac{d}{dz}\frac{z}{z+1}\bigg|_{z=1}=\frac{1}{(z+1)^2}\bigg|_{z=1}=\frac{1}{4}$. Total: $2\pi i(-1/4+1/4)=0$.
Practice
- Classify the singularities of $\frac{\sin z}{z^3}$ and find its Laurent series around $z=0$.
- Compute $\oint_{|z|=1}\frac{e^z}{z^n}\,dz$ for $n\geq 1$ using the Laurent series of $e^z$.
- State and prove Casorati-Weierstrass: near an essential singularity, $f$ is dense in $\mathbb{C}$.
- Find the residue of $\frac{\cot(\pi z)}{z^2}$ at $z=0$ and $z=n\in\mathbb{Z}\setminus\{0\}$.
Show Answer Key
1. $\frac{\sin z}{z^3}=\frac{1}{z^3}(z-z^3/6+z^5/120-\cdots)=\frac{1}{z^2}-\frac{1}{6}+\frac{z^2}{120}-\cdots$. The singularity at $z=0$ is a pole of order 2 (leading term $z^{-2}$).
2. $e^z=\sum_{k=0}^\infty z^k/k!$, so $e^z/z^n=\sum_{k=0}^\infty z^{k-n}/k!$. By the residue theorem, $\oint e^z/z^n\,dz=2\pi i\cdot\text{Res}_{z=0}=2\pi i/(n-1)!$.
3. Near an essential singularity $z_0$, for any $w\in\mathbb{C}$ and $\epsilon,\delta>0$, $\exists z$ with $0<|z-z_0|<\delta$ and $|f(z)-w|<\epsilon$. Proof by contradiction: if $|f(z)-w|\ge\epsilon$ near $z_0$, then $1/(f-w)$ is bounded and analytic, making $z_0$ a pole or removable — contradiction.
4. At $z=0$: $\cot(\pi z)/z^2$ has Laurent expansion using $\cot(\pi z)=\frac{1}{\pi z}-\frac{\pi z}{3}-\cdots$, so $\frac{\cot(\pi z)}{z^2}=\frac{1}{\pi z^3}-\frac{\pi}{3z}-\cdots$. Residue $=-\pi/3$. At $z=n\neq0$: $\text{Res}=\frac{1}{\pi n^2}$.