Editing Residue (complex analysis) (section)

==== Example 1 ====
As an example, consider the [[contour integral]]
:<math>\oint_C {e^z \over z^5}\,dz</math>
where ''C'' is some [[simple closed curve]] about 0.

Let us evaluate this integral using a standard convergence result about integration by series. We can substitute the [[Taylor series]] for <math>e^z</math> into the integrand. The integral then becomes

:<math>\oint_C {1 \over z^5}\left(1+z+{z^2 \over 2!} + {z^3\over 3!} + {z^4 \over 4!} + {z^5 \over 5!} + {z^6 \over 6!} + \cdots\right)\,dz.</math>

Let us bring the 1/''z''<sup>5</sup> factor into the series. The contour integral of the series then writes

: <math>
\begin{align}
& \oint_C \left({1 \over z^5}+{z \over z^5}+{z^2 \over 2!\;z^5} + {z^3\over 3!\;z^5} + {z^4 \over 4!\;z^5} + {z^5 \over 5!\;z^5} + {z^6 \over 6!\;z^5} + \cdots\right)\,dz \\[4pt]
= {} & \oint_C \left({1 \over\;z^5}+{1 \over\;z^4}+{1 \over 2!\;z^3} + {1\over 3!\;z^2} + {1 \over 4!\;z} + {1\over\;5!} + {z \over 6!} + \cdots\right)\,dz.
\end{align}
</math>

Since the series converges uniformly on the support of the integration path, we are allowed to exchange integration and summation.  
The series of the path integrals then collapses to a much simpler form because of the previous computation. So now the integral around ''C'' of every other term not in the form ''cz''<sup>&minus;1</sup> is zero, and the integral is reduced to

: <math>\oint_C {1 \over 4!\;z} \,dz= {1 \over 4!} \oint_C{1 \over z}\,dz={1 \over 4!}(2\pi i) = {\pi i \over 12}.</math>

The value 1/4! is the ''residue'' of ''e''<sup>''z''</sup>/''z''<sup>5</sup> at ''z'' = 0, and is denoted

: <math>\operatorname{Res}_0 {e^z \over z^5}, \text{ or } \operatorname{Res}_{z=0} {e^z \over z^5}, \text{ or } \operatorname{Res}(f,0) \text{ for } f={e^z \over z^5}.</math>