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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.

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

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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.

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