Singularities of $f(z) = \frac{z\cos(z) - z}{\sin^3(z)}$

Question

I'm having some difficulties classifying the singularities of

$$f(z) = \frac{z\cos(z) - z}{\sin^3(z)}.$$

Here's my work so far:

Using the trigonometric identity $\sin^2(z) = (1-\cos^2(z))$ is is actually quiet easy to show that the given function simplifies to

$$f(z) = \frac{-z}{\sin(z)(1 + \cos(z))}.$$

We note that

$$\lim_{z \to 0} \frac{-z}{\sin(z)(1 + \cos(z))} = \frac{-1}{2} < \infty$$

so $z_0 = 0$ is a removable singularity of $f$.

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My answer however is not complete. I have to consider the singularities when $1 + \cos(z) = 0$ but I don't know how to proceed.

Answer 1

I have to consider the singularities when $1 + \cos(z) = 0$ but I don't know how to proceed.

Assume $a,x \in \mathbb{R}$. One may recall that $$ \cos x = \cos a $$ is equivalent to $$ x = a+ 2k\pi,\quad x=-a+2k\pi, \quad k \in \mathbb{Z}, $$ then notice that $\cos \pi=-1$.

Answer 2

You do not need to do that much simplifying. Just note that $$\sin(\pi\cdot n)=0\qquad n\in\mathbb{Z}$$ Making the denominator zero, creating a singularity. However if you were to use your simplification strategy you would find that $\cos(x)=-1$ for $x=\pi+2\pi\cdot n$, $n\in\mathbb{Z}$, but since you're already having to account for the fact that $\sin(x)$ is zero for $x=\pi\cdot n$, it makes no changes to your answer.