Given a complex function
, consider the Laurent Series
![\begin{displaymath}
f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n.
\end{displaymath}](r_1288.gif) |
(1) |
Integrate term by term using a closed contour
encircling
,
The Cauchy Integral Theorem requires that the first and last terms vanish, so we have
![\begin{displaymath}
\int_\gamma f(z)\,dz = a_{-1} \int_\gamma {dz\over z-z_0}.
\end{displaymath}](r_1292.gif) |
(3) |
But we can evaluate this function (which has a Pole at
) using the Cauchy Integral Formula,
![\begin{displaymath}
f(z_0)={1\over 2\pi i} \int_\gamma {f(z)\,dz\over z-z_0}.
\end{displaymath}](r_1293.gif) |
(4) |
This equation must also hold for the constant function
, in which case it is also true that
, so
![\begin{displaymath}
1={1\over 2\pi i} \int_\gamma {dz\over z-z_0},
\end{displaymath}](r_1296.gif) |
(5) |
![\begin{displaymath}
\int_\gamma {dz\over z-z_0} = 2\pi i,
\end{displaymath}](r_1297.gif) |
(6) |
and (3) becomes
![\begin{displaymath}
\int_\gamma f(z)\,dz = 2\pi ia_{-1}.
\end{displaymath}](r_1298.gif) |
(7) |
The quantity
is known as the Residue of
at
. Generalizing to a
curve passing through multiple poles, (7) becomes
![\begin{displaymath}
\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}](r_1299.gif) |
(8) |
where
is the Winding Number and the
superscript denotes the quantity
corresponding to Pole
.
If the path does not completely encircle the Residue, take the Cauchy
Principal Value to obtain
![\begin{displaymath}
\int f(z)\,dz = (\theta_2-\theta_1)ia_{-1}.
\end{displaymath}](r_1301.gif) |
(9) |
If
has only Isolated Singularities, then
![\begin{displaymath}
\sum_{z_0^{(i)}\in \Bbb{C}^*} {a_{-1}}^{(i)} = 0.
\end{displaymath}](r_1302.gif) |
(10) |
The residues may be found without explicitly expanding into a Laurent Series as follows:
![\begin{displaymath}
f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n.
\end{displaymath}](r_1288.gif) |
(11) |
If
has a Pole of order
at
, then
for
and
. Therefore,
![\begin{displaymath}
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}](r_1306.gif) |
(12) |
![\begin{displaymath}
(z-z_0)^mf(z) = \sum_{n=0}^\infty a_{-m+n}(z-z_0)^n
\end{displaymath}](r_1307.gif) |
(13) |
Iterating,
|
|
|
(16) |
So
![\begin{displaymath}
\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}](r_1316.gif) |
(17) |
and the Residue is
![\begin{displaymath}
a_{-1} = {1\over (m-1)!} {d^{m-1}\over dz^{m-1}} [(z-z_0)^m f(z)]_{z=z_0}.
\end{displaymath}](r_1317.gif) |
(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)
© 1996-9 Eric W. Weisstein
1999-05-25