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Residue Theorem (Complex Analysis)

Given a complex function $f(z)$, consider the Laurent Series

f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n.
\end{displaymath} (1)

Integrate term by term using a closed contour $\gamma$ encircling $z_0$,

$\displaystyle \int_\gamma f(z)\,dz$ $\textstyle =$ $\displaystyle \sum_{n=-\infty}^\infty a_n \int_\gamma (z-z_0)^n\,dz$  
  $\textstyle =$ $\displaystyle \sum_{n=-\infty}^{-2} a_n \int_\gamma(z-z_0)^n\,dz + a_{-1} \int_\gamma{dz\over z-z_0}+\sum_{n=0}^\infty a_n \int_\gamma(z-z_0)^n\,dz.$ (2)

The Cauchy Integral Theorem requires that the first and last terms vanish, so we have
\int_\gamma f(z)\,dz = a_{-1} \int_\gamma {dz\over z-z_0}.
\end{displaymath} (3)

But we can evaluate this function (which has a Pole at $z_0$) using the Cauchy Integral Formula,
f(z_0)={1\over 2\pi i} \int_\gamma {f(z)\,dz\over z-z_0}.
\end{displaymath} (4)

This equation must also hold for the constant function $f(z)=1$, in which case it is also true that $f(z_0)=1$, so
1={1\over 2\pi i} \int_\gamma {dz\over z-z_0},
\end{displaymath} (5)

\int_\gamma {dz\over z-z_0} = 2\pi i,
\end{displaymath} (6)

and (3) becomes
\int_\gamma f(z)\,dz = 2\pi ia_{-1}.
\end{displaymath} (7)

The quantity $a_{-1}$ is known as the Residue of $f(z)$ at $z_0$. Generalizing to a curve passing through multiple poles, (7) becomes
\int_\gamma f(z)\,dz
= 2\pi i \sum_{i=1}^{{\rm poles\ in\ }\gamma} n(\gamma, z_0^{(i)}) a_{-1}^{(i)},
\end{displaymath} (8)

where $n$ is the Winding Number and the $(i)$ superscript denotes the quantity corresponding to Pole $i$.

If the path does not completely encircle the Residue, take the Cauchy Principal Value to obtain

\int f(z)\,dz = (\theta_2-\theta_1)ia_{-1}.
\end{displaymath} (9)

If $f$ has only Isolated Singularities, then
\sum_{z_0^{(i)}\in \Bbb{C}^*} {a_{-1}}^{(i)} = 0.
\end{displaymath} (10)

The residues may be found without explicitly expanding into a Laurent Series as follows:

f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n.
\end{displaymath} (11)

If $f(z)$ has a Pole of order $m$ at $z_0$, then $a_n=0$ for $n<-m$ and $a_{-m}\not = 0$. Therefore,
f(z) = \sum_{n=-m}^\infty a_n(z-z_0)^n = \sum_{n=0}^\infty a_{-m+n}(z-z_0)^{-m+n}
\end{displaymath} (12)

(z-z_0)^mf(z) = \sum_{n=0}^\infty a_{-m+n}(z-z_0)^n
\end{displaymath} (13)

${d\over dz} [(z-z_0)^m f(z)] = \sum_{n=0}^\infty na_{-m+n}(z-z_0)^{n-1}$
$\quad =\sum_{n=1}^\infty na_{-m+n}(z-z_0)^{n-1}$
$\quad = \sum_{n=0}^\infty (n+1)a_{-m+n+1}(z-z_0)^n$ (14)
${d^2\over dz^2} [(z-z_0)^m f(z)] = \sum_{n=0}^\infty n(n+1)a_{-m+n+1}(z-z_0)^{n-1}$
$\quad =\sum_{n=1}^\infty n(n+1)a_{-m+n+1}(z-z_0)^{n-1}$
$\quad = \sum_{n=0}^\infty (n+1)(n+2)a_{-m+n+2}(z-z_0)^n.$ (15)

${d^{m-1}\over dz^{m-1}} [(z-z_0)^m f(z)] = \sum_{n=0}^\infty (n+1)(n+2)(n+m-1) a_{n-1}(z-z_0)^n$
$= (m-1)!a_{-1} + \sum_{n=1}^\infty (n+1)(n+2)(n+m-1)a_{n-1}(z-z_0)^{n-1}.$ (16)

\lim_{z\to z_0} {d^{m-1}\over dz^{m-1}} [(z-z_0)^m f(z)] = \lim_{z\to z_0} (m-1)!a_{-1}+0 = (m-1)!a_{-1},
\end{displaymath} (17)

and the Residue is
a_{-1} = {1\over (m-1)!} {d^{m-1}\over dz^{m-1}} [(z-z_0)^m f(z)]_{z=z_0}.
\end{displaymath} (18)

This amazing theorem says that the value of a Contour Integral in the Complex Plane depends only on the properties of a few special points inside the contour.

See also Cauchy Integral Formula, Cauchy Integral Theorem, Contour Integral, Laurent Series, Pole, Residue (Complex Analysis)

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© 1996-9 Eric W. Weisstein