My sheet propose the following exercise:
Consider the function
$$
f(z)=\frac{\sin(z)}{\cos(z^3)-1}
$$
Classify the singularities and, at z = 0, calculate the residue.
It is therefore clear that there are singularities when $\cos(z^3)=1$, in particular when $z=0$. I would try, at a first glance, to solve the second part of the problem. Since I don't know at this point what kind of singularity $z_0=0$ is, I would try to write the Laurent series of $f$ near $z_0=0$. My approach is as follows: I write down the Taylor series of both the numerator and the denominator and then I try to isolate a $1/z$ term,
$$
\large
\begin{align*}
f(z) &= \frac{\sin(z)}{\cos(z^3)-1}\\
&= \frac{\left(z-\frac{z^3}{6} + \frac{z^5}{120} + \cdots\right)}{\left(1-\frac{z^6}{2}+\frac{z^{12}}{24}+\cdots\right)-1}\\
&= \frac{\left(z-\frac{z^3}{6} + \frac{z^5}{120} + \cdots\right)}{\left(-\frac{z^6}{2}+\frac{z^{12}}{24}+\cdots\right)}\\
&= \frac{z\left(1-\frac{z^2}{6}+\frac{z^4}{120}+\cdots\right)}{-\frac{z^6}2\left(1 – \frac{z^6}{12}\cdots\right)}\\
&= -\frac{2}{z^5}\frac{\left(1-\frac{z^2}{6}+\frac{z^4}{120}+\cdots\right)}{\left(1 – \frac{z^6}{12}\cdots\right)}
\end{align*}
$$
I notice that a $1/z$ term doesn't easily come out of this expression, so I force the issue a little by telling myself that the denominator term could also be kind of developed as a geometric series :
$$
\frac{1}{1-\frac{z^6}{12}+\cdots}=1 + \left(\frac{z^6}{12}+\cdots\right)+\cdots
$$
By doing this, we obtain:
$$
\begin{align*}
f(z) &= -\frac{2}{z^5}{\left(1-\frac{z^2}{6}+\frac{z^4}{120}+\cdots\right)}\left(1 + \left(\frac{z^6}{12}+\cdots\right)+\cdots\right)
\end{align*}
$$
By doing this, I can see that the second parenthesis will not be relevant for the continuation (because the first term is $1$ and beyond that we'll have positive powers of $z$). So I write :
$$
\begin{align*}
f(z) &= -\frac{2}{z^5}{\left(1-\frac{z^2}{6}+\frac{z^4}{120}+\cdots\right)}\\
&= -\frac{2}{z^5} + \frac{1}{3z^3} – \frac{1}{60z} + \mathcal O(z)
\end{align*}
$$
Now I identify the residue of $z_0=0$ as being, by definition, $-1/60$. The singularity is not essential, since there is no sum of terms in decreasing power of $z$ and the singularity is not removable since its residue is non-zero. I deduce that this is a pole of order $5$ (because it's the highest negative power of $z$).
So my question is: is my reasoning valid? It seems rather messy and fragile to me, it's à la physicist, and I'm not sure if it would work for others functions (I know that the answer I get is correct, but I don't know if I'm lucky or if the method I've used is effective). To do so, I'm surely missing some assumptions, especially when I develop the denominator into a geometric series and throw away the terms that don't interest me.
Concerning the first part of the exercise, I'm not sure how to proceed, I could start the reasoning again but it would become very cumbersome. From certain properties, I know that the other singularities are not removable, but I don't know how to classify them without carrying out an infinite number of Laurent series, as before.
Best Answer
Your reasoning is valid, but a bit too complicated. You can start in the same way, decomposing the numerator and denominator into Taylor series, but then notice that in the neighborhood of $z=0$, the value of $$\frac{1}{z^{a}+z^{b}+...}$$ is $$O\left(\frac{1}{z^{\min (a, b, ...)}}\right)$$ so you only need to consider $-\frac{1}{2}z^{6}$ in the denominator. Then simply dividing everything in the numerator up to the degree $5$ by this monomial, you obtain the same result. Perhaps this is not a very strict argument, but this is what you do in practice. What you did is essentially the same, but with more rigor and intermediate steps.