# Contour Integral from Laurent Series

Disclaimer: I'm a Physics student, not a Maths student.

I'm working on some problems involving Laurent series and the Residue theorem, and I've come across something I can't quite get my head around.

First and foremost, I had to write the Laurent series for: $$f(z)=\frac{1}{(z+1)(z-3)}$$ on:

i) The disk $$|z|<1$$

ii) The annulus $$1<|z|<3$$

I wasn't too phased by this, and obtained:

i) $$f(z)=-\frac{1}{4}\sum_{n=0}^\infty (-1)^nz^n-\frac{1}{12}\sum_{n=0}^\infty (-1)^n\left(-\frac{z}{3}\right)^n$$

ii) $$f(z)=-\frac{1}{4}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{z^n}-\frac{1}{12}\sum_{n=0}^\infty \left(\frac{z}{3}\right)^n$$

What I'm trying to do, is use this to evaluate $$\oint_C \frac{1}{(z+1)(z-3)} dz$$ on the annulus in ii)

I'm aware (from my lecture notes) that: $$\oint_C f(z) dz=2i\pi b_{-1}$$

And using my result, I found that $$b_{-1} = -\frac{1}{4}$$ Leading to: $$\oint_C \frac{1}{(z+1)(z-3)} dz = -\frac{i\pi}{2}$$

I'm not too sure if I've done this correctly in the first place, so confirmation would be good.

So I could try and verify for myself, I tried to apply the Cauchy theorem: $$\oint_C \frac{1}{(z+1)(z-3)} dz = 2i \pi \sum{Res}$$ Where I found the 2 potential residues to be: $$\pm\frac{1}{4}$$ It was obvious to me that I shouldn't use both of them, else the integral would be $$0$$. I noted that, to get the same answer as above, I should use $$-\frac{1}{4}$$

The thing I'm not too sure on, is how would I define a contour to include this pole, and exclude the other? I'm familiar with using the unit circle, but this comes with the problem of the $$-1$$ pole lying on the contour rather than inside it. Am I right in thinking that the poles are set this way, given I'm integrating over the annular region?

Any help appreciated!

You want to take a loop which is contained in $$\{z\in\Bbb C\mid1<|z|<3\}$$, like $$\gamma(t)=2e^{it}$$ ($$t\in[0,2\pi]$$). Since $$3$$ is outside the region bounded by the loop, only the residue at $$-1$$ matters. Since that residue is $$-\frac14$$, you get the same value as before.
• The loop that I mentioned was just an example. For any loop contained on the annulus that I mentioned, $-1$ is inside the region bounded by the loop, and $3$ is outside. – José Carlos Santos Apr 23 at 16:26
• Just to help clarify something; say I wanted to evaluate the integral on the disk $|z| < 3$ (as opposed to the annular region)- what predetermines the contours I can use? As selecting a loop of radius $<1$ would surely give a different (and wrong) answer? Is my choice of contour governed by the convergence of the Laurent series? I.e, even in this above example, I'd still have to pick a loop of say, radius 2, as the expansion is not valid in the region below 1? – Frankie S. Palmer Apr 23 at 16:32
• If you take a loop on hat disk, you wll have two possible answers, depending upon wether or not $-1$ is in the region bounded by the loop. – José Carlos Santos Apr 23 at 16:38
How about a circle of radius $$2$$? That will include the pole at $$z=-1$$, and exclude the one at $$z=3$$.