# Residue Theorem

Here we follow standard texts, such as Spiegel (1964)  or Levinson and Redheffer (1970). 

We consider a complex-valued function $g(z)$ which is analytic everywhere in a region ${\mathcal {R}}$ of the complex plane except possibly at a point $z=a$ .

## Case 1): $z=a$ is a simple pole

If $z=a$ is a simple pole, and $f(z)$ is analytic everywhere in ${\mathcal {R}}$ then we may write

$g(z)={\frac {f(z)}{(z-a)}}.$ By Taylor's theorem because $f(z)$ analytic, the function $f(z)$ has a Taylor expansion about $z=a$ $f(z)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(z-a)^{n}$ for every point $z$ in ${\mathcal {R}}$ . Hence the Laurent expansion may

of $g(z)$ may be formed by dividing each term of the Taylor expansion by $z-a$ . Hence the Laurent expansion of $g(z)$ is

$g(z)=\sum _{n=-1}^{\infty }{\frac {f^{(n+1)}(a)}{(n+1)!}}(z-a)^{n}={\frac {f(a)}{(z-a)}}+f^{\prime }(a)+{\frac {f^{\prime \prime }(a)}{2!}}(z-a)+...$ By the Cauchy integral formula

$f(a)={\frac {1}{2\pi i}}\int _{C}{\frac {f(z)}{(z-a)}}\;dz$ which is the coefficient of the $n=-1$ term of the Laurent expansion of $g(z)$ .

Thus, the integral over a closed contour $C$ in ${\mathcal {R}}$ circling around the point $z$ of $g(z)$ is given by

$\int _{C}g(z)\;dz=2\pi if(a)$ where $f(a)$ is the coefficient of the $n=-1$ order term of the Laurent expansion of $g(z)$ about $z=a$ ## Case 2): $z=a$ is a $M$ -th order pole.

If $z=a$ is an $M$ -th order pole, and $f(z)$ is analytic everywhere in ${\mathcal {R}}$ then we may write

$g(z)={\frac {f(z)}{(z-a)^{M}}}.$ Again, by Taylor's theorem because $f(z)$ analytic, the function $f(z)$ has a Taylor expansion about $z=a$ $f(z)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(z-a)^{n}$ for every point $z$ in ${\mathcal {R}}$ . Hence the Laurent expansion may

of $g(z)$ may be formed by dividing each term of the Taylor expansion by $(z-a)^{M}$ . Hence the Laurent expansion of $g(z)$ is

$g(z)=\sum _{n=-M}^{\infty }{\frac {f^{(n+1)}(a)}{(n+1)!}}(z-a)^{n}={\frac {f(a)}{(z-a)^{M}}}+{\frac {f^{\prime }(a)}{(z-a)^{M-1}}}+{\frac {f^{\prime \prime }(a)}{2!\;(z-a)^{M-2}}}+...+{\frac {f^{(M-1)}(a)}{(M-1)!\;(z-a)}}+{\frac {f^{(M)}(a)}{M!}}+\sum _{n=1}^{\infty }{\frac {f^{(M+n)}(a)}{(M+n)!}}(z-a)^{n}.$ As in the previous case, the n=-1 term has the integral of the function $g(z)$ is related to the $M-1$ derivative of the analytic portion $f(z)$ of $g(z).$ In this case

$\int _{C}g(z)\;dz=2\pi i{\frac {f^{(M-1)}(a)}{(M-1)!}}.$ ### The Residue $R(g(z);z=a)$ We call the quantity for a pole of order $M$ $\mathrm {Res} (g(z);z=a)={\frac {1}{(M-1)!}}\lim _{z\rightarrow a}{\frac {d^{M-1}}{dz^{M-1}}}\left((z-a)^{M}g(z)\right).$ the Residue of $g(z)$ at $z=a$ for a pole of order $M.$ ### Multiple poles at $z=\{a_{0},a_{1},a_{2},...,a_{N}\}$ Finally, if $g(z)$ has poles at $z=\{a_{0},a_{1},a_{2},...,a_{N}\}$ then by Cauchy's theorem and the Cauchy-Goursat theorem, we may replace the integral over the larger contour $C$ with the sum of integrals, each with a contour surrounding one and only one pole.

This the integral then in this case becomes

$\int _{C}g(z);dz=2\pi i\sum _{k=0}^{N}\mathrm {Res} (g(z);z=a_{k}).$ Finally, we note that for a case where the integration contours are clockwise, then there is a minus sign on the value of the integration result of Two PI i times the sum of the residues.