I encounter a question in my problem sheets, which asks to identify the type of isolated singularities of the following function: $$\frac{1}{\sin z-\sin 2z}$$
Firstly, by trig identities, I can rewrite the denominator as $-2\cos\frac{3z}{2}\,\sin\frac{z}{2}$. Therefore, the possible singularities are $\frac{\pi}{3}(2k+1)$ and $2k\pi$.
However, I have no ideas how to classify these singularities into either one of: pole, removable singularity, or essential singularity.
Personally, I think it is highly possible to be a pole, even though I don't know how to prove this. So the question is about working the order of the pole.
Many thanks for any help.
Best Answer
Hint:
The poles are the solutions of the equation $$\sin z=\sin 2z\iff \begin{cases} 2z\equiv z \\2z\equiv \pi-z \end{cases}\mod 2\pi\iff \begin{cases} z\equiv 0 \\3z\equiv \pi \end{cases}\mod 2\pi\iff \begin{cases} z\equiv 0 \mod 2\pi\\z\equiv \frac \pi3\mod \frac{2\pi}3 \end{cases} $$ You can check that none of these poles is a root of the derivative $$(\sin z-\sin 2z)'=\cos z-2\cos 2z,$$ so that the poles are simple poles.