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!
Best Answer
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.