Editing Residue (complex analysis) (section)

== Examples<!-- Needs simple pole and higher order pole residue examples below --> ==

=== Residue from series expansion ===

==== 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>

==== Example 2 ====
As a second example, consider calculating the residues at the singularities of the function<math display="block">f(z) = {\sin z \over z^2-z}</math>which may be used to calculate certain contour integrals. This function appears to have a singularity at ''z'' = 0, but if one factorizes the denominator and thus writes the function as<math display="block">f(z) = {\sin z \over z(z - 1)}</math>it is apparent that the singularity at ''z'' = 0 is a [[removable singularity]] and then the residue at ''z'' = 0 is therefore 0.
The only other singularity is at ''z'' = 1. Recall the expression for the Taylor series for a function ''g''(''z'') about ''z'' = ''a'':<math display="block"> g(z) = g(a) + g'(a)(z-a) + {g''(a)(z-a)^2 \over 2!} + {g'''(a)(z-a)^3 \over 3!}+ \cdots</math>So, for ''g''(''z'') = sin&nbsp;''z'' and ''a'' = 1 we have<math display="block"> \sin z = \sin 1 + (\cos 1)(z-1)+{-(\sin 1)(z-1)^2 \over 2!} + {-(\cos 1)(z-1)^3 \over 3!} + \cdots.</math>and for ''g''(''z'') = 1/''z'' and ''a'' = 1 we have<math display="block"> \frac{1}{z} = \frac1 {(z - 1) + 1} = 1 - (z - 1) + (z - 1)^2 - (z - 1)^3 + \cdots.</math>Multiplying those two series and introducing 1/(''z''&nbsp;−&nbsp;1) gives us<math display="block"> \frac{\sin z} {z(z - 1)} = {\sin 1 \over z-1} + (\cos 1 - \sin 1) + (z-1) \left(-\frac{\sin 1}{2!} - \cos1 + \sin 1\right) + \cdots.</math>So the residue of ''f''(''z'') at ''z'' = 1 is sin&nbsp;1.

==== Example 3 ====

The next example shows that, computing a residue by series expansion, a major role is played by the [[Formal series#The Lagrange inversion formula|Lagrange inversion theorem]]. Let<math display="block"> u(z) := \sum_{k\geq 1}u_k z^k</math>be an [[entire function]], and let<math display="block">v(z) := \sum_{k\geq 1}v_k z^k</math>with positive radius of convergence, and with <math display="inline"> v_1 \neq 0</math>. So <math display="inline"> v(z)</math> has a local inverse <math display="inline"> V(z)</math> at 0, and <math display="inline"> u(1/V(z))</math> is [[meromorphic]] at 0. Then we have:<math display="block">\operatorname{Res}_0 \big(u(1/V(z))\big) = \sum_{k=0}^\infty ku_k v_k. </math>Indeed,<math display="block">\operatorname{Res}_0\big(u(1/V(z))\big) = \operatorname{Res}_0 \left(\sum_{k\geq 1} u_k V(z)^{-k}\right) = \sum_{k\geq 1} u_k \operatorname{Res}_0 \big(V(z)^{-k}\big)</math>because the first series converges uniformly on any small circle around 0. Using the Lagrange  inversion theorem<math display="block">\operatorname{Res}_0 \big(V(z)^{-k}\big) = kv_k,</math>and we get the above expression. For example, if <math>u(z) = z + z^2</math> and also <math>v(z) = z + z^2</math>, then<math display="block">V(z) = \frac{2z}{1 + \sqrt{1 + 4z}}</math>and<math display="block">u(1/V(z)) = \frac{1 + \sqrt{1 + 4z}}{2z} + \frac{1 + 2z + \sqrt{1 + 4z}}{2z^2}.</math>The first term contributes 1 to the residue, and the second term contributes 2 since it is asymptotic to <math>1/z^2 + 2/z</math>.


Note that, with the corresponding stronger symmetric assumptions on <math display="inline"> u(z)</math> and <math display="inline"> v(z)</math>, it also follows<math display="block">\operatorname{Res}_0 \left(u(1/V)\right) = \operatorname{Res}_0\left(v(1/U)\right),</math>where <math display="inline"> U(z)</math> is a local inverse of <math display="inline"> u(z)</math> at&nbsp;0.