<s>[INST] Power series of z/sin(z)? So I need to compute the coefficient of the $z^4$ in the power series of $\frac{z}{\sin z}$.
I tried differentiating the function and obtaining coefficients like in Taylor's expansions but had a really hard time. In general I'm finding it extremely hard to obtain the power series of various complex functions, especially with something like $\sin z$ in the denominator. Any tips / tricks?
 [/INST] Here is a simple method that works well to find the first few terms in the power-series.
We first expand the denominator in a power-series around $z=0$: $$\frac{z}{\sin(z)} = \frac{1}{1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \ldots}$$
This has the form of a geometrical series $\frac{1}{1-x} = 1+x + x^2 + \ldots$ for $x = \frac{z^2}{3!} - \frac{z^4}{5!} + \mathcal{O}(z^6)$. The expansion above holds as long as $|x|<1$ which is the case for all $z$ sufficiently close to $z=0$ (which is all we need).
Expanding the different terms $x^k$ is a power-series in $z$ we find
$$\begin{align}x^2 &= \left(\frac{z^2}{3!}\right)^2 - 2\frac{z^2}{3!}\frac{z^4}{5!} + \mathcal{O}(z^8)\\x^3 &= \left(\frac{z^2}{3!}\right)^3 + \mathcal{O}(z^8)\end{align}$$ 
The term $x^n$ will only give rise to terms of order $z^{2n}$ and larger so to get the series to order $z^6$ we can stop at $x^3$. Adding up all the terms above gives 
$$\frac{z}{\sin(z)} = 1 + \frac{z^2}{6} + \frac{7z^4}{360} + \frac{31z^6}{15120} + \mathcal{O}(z^8)$$
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