How to find the Laurent series expansion of a complex function?

**Taylor vs Laurent:** - **Taylor series:** Only non-negative powers. Valid in a disk $|z - z_0| < R$ where $f$ is analytic. - **Laurent series:** Includes negative powers. Valid in an annulus $r < |z - z_0| < R$ where $f$ is analytic (possibly with singularities inside). **Example 1: $f(z) = \frac{1}{z(z-1)}$ in $0 < |z| < 1$.** Use partial fractions: $\frac{1}{z(z-1)} = -\frac{1}{z} + \frac{1}{z-1}$. For $|z| < 1$: $\frac{1}{z-1} = -\frac{1}{1-z} = -\sum_{n=0}^{\infty} z^n$. Thus: $$f(z) = -\frac{1}{z} - \sum_{n=0}^{\infty} z^n = -\frac{1}{z} - 1 - z - z^2 - z^3 - \cdots$$ The negative power term $-1/z$ tells us $z = 0$ is a simple pole. **Example 2: $f(z) = e^{1/z}$ about $z = 0$.** Use the Taylor series for $e^w$ with $w = 1/z$: $$e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n!} \left(\frac{1}{z}\ ight)^n = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \cdots$$ This has **infinitely many** negative power terms, so $z = 0$ is an **essential singularity**. **Classification via Laurent Series:** - **Removable singularity:** No negative powers (e.g., $\sin z/z$) - **Pole of order $m$:** Finite negative powers, highest is $1/(z-z_0)^m$ - **Essential singularity:** Infinitely many negative powers (e.g., $e^{1/z}$) **Annulus of convergence:** Determined by the locations of singularities. For $f(z) = \frac{1}{z(z-1)}$, singularities at $z = 0$ and $z = 1$, so there are three regions: $|z| < 1$, $1 < |z|$, and — no, actually: $0 < |z| < 1$ (as above) and $1 < |z|$ (different expansion).