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\documentclass{article}
\begin{document}
\noindent
Laurent expansion using $c_n = \frac{1}{2\pi i}
\oint (\zeta-a)^{-n-1}f(\zeta)\,d\zeta$, for every function $f(z)$ the
following representation is valid ($n=0$, $\pm1$, $\pm2$, \ldots)
\[ f(x) = \sum_{n=-\infty}^{+\infty} c_n(z-a)^n
= \left\{\begin{array}{r}
c_0 + c_1(z-a) + c_2(z-a)^2 +\cdots+ c_n(z-a)^n+\cdots\\
\mbox{}+c_{-1}(z-a)^{-1} + c_{-2}(z-a)^{-2}+\cdots\\
\mbox{}+c_{-n}(z-a)^{-n}+\cdots \end{array}\right. \]
\end{document}
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