In working through problems for my complex analysis class, I've come across a couple of problems about Laurent series and residue where the function to be expanded, $g(z)$, is of the form $f(f(z))$ where $f$ is some function like $\cos(z)$ or $\sin(z)$ that we already know the expansion for.
I've tried to construct the expansions from what I know about trigonometric expansions, but it just seems really messy and I can never quite get a clean expression for the expansion and in the case of residues it feels like there has to be an easier way.
So my question is, given something similar to, \begin{align} g(z) = \frac{\sin(\sin(\frac{1}{z}))}{z^5} \end{align} how do I go about constructing the expansion and/or calculating the residue at the singularity?
It seems to me that in general, it would be rather messy. For instance, for the composition of just two power series the general form is not necessarily beautiful... there is not even a nice form for the composition of polynomials.
However, when $f$'s Laurent series only has finitely many $a_n \neq 0$ for $n < 0$, it seems that there is an easier method of doing this. If $N$ is the least $n$ such that $a_n \neq 0$, then you can write express your function as $f(1/z^N)$, where $f$ is holomorphic at $0$. Then you can calculate the relevant coefficient like you would for a normal power series starting at $0$.
For instance, in your function, there is no residue since the greatest power term that you will have is $1/z^5$, so I suppose that you were interested in instead $z^5\sin(\sin(1/z))$.