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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): z = \{ a_0, a_1, a_2, ... , a_N\}

Finally, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): g(z) has poles at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): C with the sum of integrals, each with a contour surrounding one and only one pole.

This the integral then in this case becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \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.