Editing Residue (complex analysis) (section)

===Simple poles===
If ''c'' is a [[simple pole]] of ''f'', the residue of ''f'' is given by:

:<math>\operatorname{Res}(f,c)=\lim_{z\to c}(z-c)f(z).</math>

If that limit does not exist, then ''f'' instead has an essential singularity at ''c''. If the limit is 0, then ''f'' is either analytic at ''c'' or has a removable singularity there. If the limit is equal to infinity, then the order of the pole is higher than 1.

It may be that the function ''f'' can be expressed as a quotient of two functions, <math>f(z)=\frac{g(z)}{h(z)}</math>, where ''g'' and ''h'' are [[holomorphic function]]s in a [[Neighbourhood (mathematics)|neighbourhood]] of ''c'', with ''h(c)'' = 0 and&nbsp;''h'(c)''&nbsp;≠&nbsp;0. In such a case, [[L'Hôpital's rule]] can be used to simplify the above formula to:

: <math>
\begin{align}
\operatorname{Res}(f,c) & =\lim_{z\to c}(z-c)f(z) = \lim_{z\to c}\frac{z g(z) - cg(z)}{h(z)} \\[4pt]
& = \lim_{z\to c}\frac{g(z) + z g'(z) - cg'(z)}{h'(z)} = \frac{g(c)}{h'(c)}.
\end{align}
</math>