Classifying a singularity $f(z)=\frac{\sin z}{\cos(z^3)-1}$ $$f(z)=\frac{\sin z}{\cos(z^3)-1}$$
I need to classify the singularity at $z=0$ and calculate the residue.
Apparently the method of evaluating this is through series, but that seems strange to me since I have only encountered apparently trivial cases through cauchy's theorem. What is the easiest approach to calculating the residue?
 A: As you mention, we can understand this through series. (One could also proceed using the formula that Arthur mentions in his comment).
Firstly, we know that $\sin z = z - z^3/6 + \dots$ and $\cos(z^3) - 1 = (1-1) - z^6/2 + \ldots = -z^6/2 + \dots$. So the ratio looks like
$$ \frac{z - z^3/6}{-z^6/2} = \frac{1 - z^2/6}{-z^5/2}.$$
We've omitted higher order terms because they do not contribute here. This tells us that we have a pole of order $5$.
You also ask how to compute the residue, which is the coefficient of the degree $-1$ term. Let's show the let's use series method of getting the residue.
We now know that
$$ \frac{\sin z}{\cos z^3 - 1} = \frac{a_{-5}}{z^5} + \frac{a_{-4}}{z^4} + \dots + a_0 + a_1z + \dots$$
This also means that
$$ \sin z = (\cos z^3 - 1)\left(\frac{a_{-5}}{z^5} + \frac{a_{-4}}{z^4} + \dots + a_0 + a_1z + \dots\right).$$
We can now compare series expansions to arrive at the answer. In particular,
$$ z - z^3/6 + z^5/120 + \dots = (-z^6/2 + z^{12}/6 + \dots)\left(\frac{a_{-5}}{z^5} + \frac{a_{-4}}{z^4} + \dots + a_0 + a_1z + \dots\right).$$
We know that $a_{-5} = -2$ from above. Notice that $a_{-4} = a_{-2} = a_0 = 0$, which we can see by noticing that the products with $-z^6$ have degrees that don't appear on the left or by noticing that the left side is an odd function and $\cos z^3 - 1$ is even, so the Laurent expansion must be an odd function.
So how about $a_{-1}$? Notice that the only way to multiply two terms on the right to get a term like $z^5$ is to multiply $-z^6/2$ and $a_{-1}/z$. So by comparing coefficients, we see that
$$ a_{-1} \cdot \frac{-1}{2} = \frac{1}{120},$$
so that
$$a_{-1} = -\frac{1}{60}.$$
So this is the residue. *As an aside, notice that we didn't even have to compute the $a_{-3}$ term. This sort of thing can happen when we have really sparse series, like $\cos z^3 - 1$. In these cases, we can very quickly compute residues with series, especially when many things can be done quickly in your head, as in this case.*$\diamondsuit$
