# Residue Theorem

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Here we follow standard texts, such as Spiegel (1964) [1] or Levinson and Redheffer (1970). [2]

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

## Case 1): ${\displaystyle z=a}$ is a simple pole

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

${\displaystyle g(z)={\frac {f(z)}{(z-a)}}.}$

By Taylor's theorem because ${\displaystyle f(z)}$ analytic, the function ${\displaystyle f(z)}$ has a Taylor expansion about ${\displaystyle z=a}$

${\displaystyle f(z)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(z-a)^{n}}$

for every point ${\displaystyle z}$ in ${\displaystyle {\mathcal {R}}}$ . Hence the Laurent expansion may

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

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

By the Cauchy integral formula

${\displaystyle f(a)={\frac {1}{2\pi i}}\int _{C}{\frac {f(z)}{(z-a)}}\;dz}$

which is the coefficient of the ${\displaystyle n=-1}$ term of the Laurent expansion of ${\displaystyle g(z)}$.

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

${\displaystyle \int _{C}g(z)\;dz=2\pi if(a)}$

where ${\displaystyle f(a)}$ is the coefficient of the ${\displaystyle n=-1}$ order term of the Laurent expansion of ${\displaystyle g(z)}$ about ${\displaystyle z=a}$

## Case 2): ${\displaystyle z=a}$ is a ${\displaystyle M}$-th order pole.

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

${\displaystyle g(z)={\frac {f(z)}{(z-a)^{M}}}.}$

Again, by Taylor's theorem because ${\displaystyle f(z)}$ analytic, the function ${\displaystyle f(z)}$ has a Taylor expansion about ${\displaystyle z=a}$

${\displaystyle f(z)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(z-a)^{n}}$

for every point ${\displaystyle z}$ in ${\displaystyle {\mathcal {R}}}$ . Hence the Laurent expansion may

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

${\displaystyle g(z)=\sum _{n=-M}^{\infty }{\frac {f^{(n+1)}(a)}{(n+1)!}}(z-a)^{n}={\frac {f(a)}{(z-a)^{M}}}+{\frac {f^{\prime }(a)}{(z-a)^{M-1}}}+{\frac {f^{\prime \prime }(a)}{2!\;(z-a)^{M-2}}}+...+{\frac {f^{(M-1)}(a)}{(M-1)!\;(z-a)}}+{\frac {f^{(M)}(a)}{M!}}+\sum _{n=1}^{\infty }{\frac {f^{(M+n)}(a)}{(M+n)!}}(z-a)^{n}.}$

As in the previous case, the n=-1 term has the integral of the function ${\displaystyle g(z)}$ is related to the ${\displaystyle M-1}$ derivative of the analytic portion ${\displaystyle f(z)}$ of ${\displaystyle g(z).}$ In this case

${\displaystyle \int _{C}g(z)\;dz=2\pi i{\frac {f^{(M-1)}(a)}{(M-1)!}}.}$

### The Residue ${\displaystyle R(g(z);z=a)}$

We call the quantity for a pole of order ${\displaystyle M}$

${\displaystyle \mathrm {Res} (g(z);z=a)={\frac {1}{(M-1)!}}\lim _{z\rightarrow a}{\frac {d^{M-1}}{dz^{M-1}}}\left((z-a)^{M}g(z)\right).}$

the Residue of ${\displaystyle g(z)}$ at ${\displaystyle z=a}$ for a pole of order ${\displaystyle M.}$

### Multiple poles at ${\displaystyle z=\{a_{0},a_{1},a_{2},...,a_{N}\}}$

Finally, if ${\displaystyle g(z)}$ has poles at ${\displaystyle z=\{a_{0},a_{1},a_{2},...,a_{N}\}}$ then by Cauchy's theorem and the Cauchy-Goursat theorem, we may replace the integral over the larger contour ${\displaystyle C}$ with the sum of integrals, each with a contour surrounding one and only one pole.

This the integral then in this case becomes

${\displaystyle \int _{C}g(z);dz=2\pi i\sum _{k=0}^{N}\mathrm {Res} (g(z);z=a_{k}).}$

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.

## References

1. Spiegel, Murray R. "Theory and problems of complex variables, with an introduction to Conformal Mapping and its applications." Schaum's outline series (1964).
2. Levinson, Norman, and Raymond M. Redheffer. "Complex variables." (1970), Holden-Day, New York.