Residue Theorem (Complex Analysis)

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

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

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

$$\int_\gamma f(z)\,dz = \sum_{n=-\infty}^\infty a_n \int_\gamma (z-z_0)^n\,dz$$

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

The Cauchy Integral Theorem requires that the first and last terms vanish, so we have

$$\int_\gamma f(z)\,dz = a_{-1} \int_\gamma {dz\over z-z_0}.$$ (3)

But we can evaluate this function (which has a Pole at $z_0$) using the Cauchy Integral Formula,

$$f(z_0)={1\over 2\pi i} \int_\gamma {f(z)\,dz\over z-z_0}.$$ (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

$$1={1\over 2\pi i} \int_\gamma {dz\over z-z_0},$$ (5)

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

and (3) becomes

$$\int_\gamma f(z)\,dz = 2\pi ia_{-1}.$$ (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

$$\int_\gamma f(z)\,dz = 2\pi i \sum_{i=1}^{{\rm poles\ in\ }\gamma} n(\gamma, z_0^{(i)}) a_{-1}^{(i)},$$ (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

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

If $f$ has only Isolated Singularities, then

$$\sum_{z_0^{(i)}\in \Bbb{C}^*} {a_{-1}}^{(i)} = 0.$$ (10)

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

$$f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n.$$ (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,

$$f(z) = \sum_{n=-m}^\infty a_n(z-z_0)^n = \sum_{n=0}^\infty a_{-m+n}(z-z_0)^{-m+n}$$ (12)

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

Iterating,

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

So

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

and the Residue is

$$a_{-1} = {1\over (m-1)!} {d^{m-1}\over dz^{m-1}} [(z-z_0)^m f(z)]_{z=z_0}.$$ (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)