# 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.

• That makes sense. But how would I illustrate that I've taken this loop as opposed to any other, since the integral would look the same? Would I literally just have to state it? Apr 23 '20 at 16:24
• 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. Apr 23 '20 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? Apr 23 '20 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. Apr 23 '20 at 16:38
• And is there any indicator as to which one would be correct? Or are both right? Apr 23 '20 at 16:38

How about a circle of radius $$2$$? That will include the pole at $$z=-1$$, and exclude the one at $$z=3$$.

• As I've also commented below- I think the thing I'm a tad confused with is how to illustrate I've done this? Like, what's to stop me choosing any other loop? Or is the choice completely arbitrary, as long as it's on the annulus? Apr 23 '20 at 16:25
• Yes, by the residue theorem, you can choose any such loop.
– user403337
Apr 23 '20 at 16:26