# What exactly is a residue - what are its applications[complex analysis]

1. Apr 27, 2007

### trickae

In https://www.amazon.com/Fundamentals...6816057?ie=UTF8&s=books&qid=1177661695&sr=8-1 - residues is introduced as an exercise at the end of a chapter and thats it! (or it may resurface in a later chapter),

My question is that saff and snider looks at it as the numerator of the partial fraction exapansion of a polynomail fraction.

But in Schaums series we have a nice little function like this:

Code (Text):

a = lim   1/(k-1)! . [color=red](d^(k-1) /dz^(k-1)) [/color] {(z-a)^k f(z)}
z->a

where the term in red is the differential operator and the order is determined by k-1

so whats this used for? which method is right? why choose one method over the other? And what is it beside the sum of all the residues at the singularities = the integral of the function that contains it - i.e. f(z) ?

sorry if this is a silly question.

Last edited by a moderator: Apr 22, 2017
2. Apr 27, 2007

### HallsofIvy

Staff Emeritus
If you integrate zn on a closed curve around 0 what do you get?
More generally, if you integrate (z-z0)n on a close curve containing z0 what do you get?

That's easy to answer. Since (z-z0)n is analytic everywhere except at z0, we get the same for all such curves so we can look at the circle, of radius 1 around z0, taking $z= z_0+ e^{i\theta}$ so that $(z-z_0)^n= e^{ni\theta}$ and $dz= ie^{i\theta}d\theta$:
$$\int (z-z_0)^n dz= \int_0^{2\pi}ie^{(n+1)i\theta}d\theta$$
As long as n is not equal to -1, that is
$$\frac{-i}{n+1}e^{(n+1)i\theta}$$
evaluated at $\theta= 0$ and $\theta= 2\pi$. But $e^{(n+1)i\theta}$ is 0 at both ends so the integral is 0.

If n= -1, then $e^{(n+1)i\theta}= e^0= 1$ so the integral is just
$$\int_0^{2\pi} d\theta= 2\pi i[/itex]. Now, suppose f(z) is a function having a "pole of order n" at z= z0. That means it can be written as a power series with powers of z down to -n: a "Laurent series". [tex]f(z)= a_{-n}(z-z_0)^{-n}+ \cdot\cdot\cdot+ a_{-1}(z-z_0)^{-1}+ \cdot\cdot\cdot$$
Integrating that, term by term, on a contour containing z0, every term gives 0 except the (z-z0)-1 term. That term gives $2\pi i a_{-1}$ so the integral is that.

The "residue" of f(z) at z0 is precisely the coefficient of (z-z0)-1 in a Laurent series expansion of f(z). It can be found using a formula similar to the formula for Taylor's series coefficients- That's the formula Schaum's outline is giving you. If f(z) is analytic inside a countour except for some points at which it has poles, the integral of f around that contour is just $2\pi i$ times the sum of the residues at each of those poles.

3. Apr 28, 2007

### trickae

thanks a tonne halls of ivy

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