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\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)
\[ f(x) = \sum_{n=-\infty}^{+\infty} c_n(z-a)^n
	= \left\{\begin{array}{r}
	  c_0 + c_1(z-a) + c_2(z-a)^2 +\cdots+ c_n(z-a)^n+\cdots\\
	  \mbox{}+c_{-1}(z-a)^{-1} + c_{-2}(z-a)^{-2}+\cdots\\
	  \mbox{}+c_{-n}(z-a)^{-n}+\cdots \end{array}\right. \]
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