Classify singularities in the extended complex plane

Question

I'm trying to find and classify the singularities of the function:

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

in the extended complex plane.

I've been trying to expand the Laurent series at $\frac{\pi}{2} +2k\pi$ and at $\infty$ but without any success...

Answer 1

For finding singularities of $f(z)=\dfrac{1}{1-\sin(z)}$ it is sufficient to see when does the denominator zero. $\sin z=1$, and we know zeros of $\sin$ function are real, so singularities are $z=2k\pi+\dfrac{\pi}{2}$. For the type of singularities we consider that they are poles of order $2$ since $$\operatorname*{Res}_{z=2k\pi+\frac{\pi}{2}}\dfrac{1}{1-\sin(z)}=0$$ Edit. Let $w=z-(2k\pi+\dfrac{\pi}{2})$ then $$\dfrac{1}{1-\sin z}=\dfrac{1}{1-\cos w}=\dfrac{1}{\frac{1}{2!}w^2-\frac{1}{4!}w^4+\cdots}=\dfrac{2}{w^2}+\dfrac16+\dfrac{w^2}{120}+\cdots$$

Answer 2

Hint:

Note that the singularities of $f(z)$ are those $z$ for which $$1- \sin z =0 \implies \sin z =1 \implies z = \,?$$

Answer 3

This function has poles at $z_0 = \frac\pi2 + 2n\pi$. One way to find the order of the poles is to find the order of the zeroes of $$ g(z) = 1-\sin z $$

We know that $g(z_0) = 0$ and $g'(z_0) = -\cos(\frac\pi2+2n\pi) = 0$, while $g''(z_0) = \sin(\frac\pi2+2n\pi) = 1 \ne 0$. Therefore these are second-order zeroes of $g(z)$ and consequently, second-order poles of $f(z)$

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It isn't necessary to use the Laurent series, but if you want to check the lowest power, substitute $w = z - \frac\pi2 - 2n\pi$

$$ \begin{align} \frac{1}{1-\sin \left(w + \frac\pi2 + 2n\pi \right)} &= \frac{1}{1 - \cos w} \\ &= \frac{1}{1 - \left(1 - \dfrac{w^2}{2} + \dfrac{w^4}{4!} + \dots \right)} \\ &= \frac{2}{w^2}\frac{1}{1-\dfrac{2w^2}{4!} + \cdots} \\ &= \frac{2}{w^2}\left(1 + \frac{2w^2}{4!} + \dots \right) \end{align} $$