coefficients of Laurent series

Suppose that $f$ is analytic in the annulus  ,  where $R_1$ may be 0 and $R_2$ may be $\mathrm{\infty }$.  Then the coefficients of the Laurent series (http://planetmath.org/LaurentSeries)

$$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{(z-a)}^n$$

of $f$ can be obtained from

$c_n={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\oint }_{\gamma }}{\displaystyle \frac{f(t)}{{(t-a)}^{n+1}}}dt\mathit{\hspace{1em}}(n=0,\pm 1,\pm 2,\mathrm{\dots }),$(1)

where the path (http://planetmath.org/ContourIntegral) $\gamma$ goes anticlockwise once around the point  $z=a$  within the annulus.  Especially, the residue of $f$ in the point $a$ is

$c_{-1}={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\oint }_{\gamma }}f(t)𝑑t.$(2)

Remark.  Usually, the Laurent series of a function, i.e. the coefficients $c_n$, are not determined by using the integral formula (1), but directly from known series .  Often it is sufficient to know the value of $c_{-1}$ or the residue, which is used to compute integrals (see the Cauchy residue theorem —  cf. (2)).  There is also the usable

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

$$\mathrm{Res}(f;a)=r.$$

Examples

1. Let  $f(z):=\frac{1}{\sin z}$,  and  $a=0$.  Using the Taylor series of the complex sine we obtain

   $$\underset{z\to 0}{lim}z\frac{1}{\sin z}=\underset{z\to 0}{lim}\frac{1}{1-\frac{z^2}{3!}+-\mathrm{\dots }}=\mathrm{\hspace{0.33em}1},$$

   whence  $\mathrm{Res}(\frac{1}{\sin z};\mathrm{\hspace{0.17em}0})=1$.  Thus we can write

   $${\oint }_{\gamma }\frac{dz}{\sin z}=\mathrm{\hspace{0.33em}2}\pi i,$$

   where the 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

   $$e^{\frac{1}{z}}\equiv \mathrm{\hspace{0.33em}1}+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\mathrm{\dots }$$

   which shows that  $\mathrm{Res}(e^{\frac{1}{z}};\mathrm{\hspace{0.17em}0})=1.$