Problem: The function
$f(z) = \frac{(z^{2}-2z-3)^{2}}{\cos(\pi z)+1}$
has an isolated singularity at $z_0=1$.
a) Find the principal (singular) part of the Laurent expansion of $f$ in a punctured neighbourhood of $z_0=1$.
b) In which region does the Laurent expansion converge?
My thoughts: Usually I am given the region in which the Laurent expansion converges but now I have to determine that myself. Will the Laurent series still be uniquely determined? If so, I have an idea about how to proceed: to manipulate the function so that we have $z-1$ in place of $z$, find Taylor series for every $z-1$ and then just divide when you have the expansion in the top and the bottom. I am not sure if this is a good idea or not and if so I do not know how to change $cos(\pi z)$ into something I can work with.
All input appreciated! I am studying for an exam in complex analysis, this is a problem from an old exam.
Best Answer
The Laurent series in any given annulus is uniquely determined. (You should look up a proof of this. It's important.) In this case, you have convergence in a punctured disk. To determine this disk, look at the singularities of the function and note the distance from 1 to the nearest singularity. (If you return to the proof of the existence of Laurent series, you will see that the radius of convergence will be the largest possible punctured disk around with point that doesn't contain another singularity.)
The hard part about finding the Laurent series here is inverting the denominator. If you write out the power series for the denominator, you will see that it is
$$(z-1)^2(a+b(z-1)+c(z-1)^2+\dots)$$
where $a,b,c$ are some constants. So you can factor out the $\frac{1}{(z-1)^2}$ term and use the normal procedure for inverting power series with a nonzero constant term. You can use your idea (writing everything in terms of $z-1$) to take care of the numerator.