# Determining the order of the poles of the function $\frac{1}{\sin z-\sin 2z}$

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.

• Can you determine the orders of the zeros of the denominators? If they're simple zeros, fr example, then the function has simple poles. – saulspatz Mar 23 '19 at 13:48
• @saulspatz Unfortunately, I can't :( – Jamie Carr Mar 23 '19 at 13:49
• If it's a multiple zero, then the derivative vanishes also. – saulspatz Mar 23 '19 at 13:50

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.
The function can be rewritten $$f(z)=\frac{1}{\sin z}\frac{1}{1-2\cos z}$$ Note that $$\sin z$$ and $$1-2\cos z$$ don't vanish at the same points, so it's sufficient to study the singularities of $$1/\!\sin z$$ and of $$1/(1-2\cos z)$$.
All of them are simple poles. Since $$1-2\cos z=A\sin(z+\varphi)$$ for some $$\phi$$, it's sufficient to show that the poles of $$1/\!\sin z$$ are simple and I guess you're able to do it.