coefficients of Laurent seriesCoefficientsOfLaurentSeries2005-05-30 17:08:562009-06-05 18:26:15ResultLaurent series% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}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 \PMlinkname{Laurent series}{LaurentSeries} \PMlinkescapetext{expansion}
$$\sum_{n = -\infty}^\infty c_n (z-a)^n$$
of $f$ can be obtained from
\begin{align}
c_n \;=\; \frac{1}{2\pi i}\oint_{\gamma} \frac{f(t)}{(t-a)^{n+1}}\,dt
\quad (n = 0,\,\pm 1,\,\pm 2,\,\ldots),
\end{align}
where the \PMlinkname{path}{ContourIntegral} $\gamma$ goes anticlockwise once around the point \,$z = a$\, within the annulus.\, Especially, the residue of $f$ in the point $a$ is
\begin{align}
c_{-1} \;=\; \frac{1}{2\pi i}\oint_{\gamma} f(t)\,dt.
\end{align}
\textbf{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 \PMlinkescapetext{expansions}.\, 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
\textbf{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 \PMlinkescapetext{order} 1 for the function $f$ and
$$\operatorname{Res}(f;\,a) \;=\; r.$$
\textbf{Examples}
\begin{enumerate}
\item 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 \PMlinkescapetext{path} must be chosen such that it encloses only the pole $0$ of
$\frac{1}{\sin{z}}$.
\item 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\, $\operatorname{Res}(e^{\frac{1}{z}};\,0) = 1.$
\end{enumerate}