\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}