Suppose that $f$ is analytic in the annulus $\{z\in\mathbb{C}\,\vdots\,\, R_1 < |z-a| < R_2 \}$ , where $R_1$ may be 0 and $R_2$ may be $\infty$ . Then the coefficients of the Laurent series expansion $$\sum_{n = -\infty}^\infty c_n (z-a)^n$$ of $f$ can be obtained from

(1)

(2)

Rule. In the case that the limit $\displaystyle\lim_{z\to a}(z-a)f(z)$ exists and has a non-zero value $r$ , the point $z = a$ is a pole of the order 1 for the function $f$ and $$\operatorname{Res}(f;\,a) \;=\; r.$$

Examples

1. Let $f(z) := \frac{1}{\sin{z}}$ , and $a = 0$ . Using the Taylor series of the complex sine we obtain $$\lim_{z\to 0}z\frac{1}{\sin{z}} \;=\; \lim_{z\to 0}\frac{1}{1-\frac{z^2}{3!}+-\ldots} \;=\; 1,$$ whence $\operatorname{Res}(\frac{1}{\sin{z}};\,0) = 1$ . Thus we can write $$\oint_{\gamma}\frac{dz}{\sin{z}} \;=\; 2\pi i,$$ where the path must be chosen such that it encloses only the pole $0$ of $\frac{1}{\sin{z}}$ .
2. The Taylor series of the complex exponential function gives the Laurent series $$e^{\frac{1}{z}} \;\equiv\; 1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\ldots$$ which shows that $\operatorname{Res}(e^{\frac{1}{z}};\,0) = 1.$