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I am having trouble with a specific problem actually. I have a function $f(z) = \frac{1}{z\cdot \sin{z}}$

Now I want to find the residues of this. The Laurent series expanded about $0$ shows that $0$ is a pole of order $2$. The expansion looks something like this $\frac{1}{z^2}+\frac{1}{6} + \frac{7z^2}{360} + \cdots $

so since the first coefficient of $z$ is just zero, the residue of this function is $0$.

BUT I want to know why zero is the ONLY pole. Clearly $2\pi$ is a singularity point. Then when you expand about $2\pi$ you get the following expansion $\frac{1}{2\pi (z - 2\pi)} - \frac{1}{4\pi^2} + \frac{(3+2\pi^2)(z-2\pi)}{24\pi^3} + \cdots $

Again, it looks to me that the first negative power of $z$ has the coefficient $\frac{1}{2\pi}$.

So why is it that when I type in "poles of function 1/(z*sin(z))" wolfram only identifies 0 as the pole. If I type in "poles of function 1/(sin(z))" then it identifies the poles as $n\pi$. Furthermore if you type in "residues of 1/(z*sin(z))" it only identifies 0 as a residue when we just saw above that $\frac{1}{2\pi}$ is also a residue. Whats even more weird is that if you type in "residues of 1/(z*sin(z)) at 2pi" it does give the right residue. Weird.

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    Next you should ask Wolfram, since we do not know its reasons for that error.2012-05-21

2 Answers 2

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Rather than expanding the function $f(z)$ around the point $z=2\pi$, let's rearrange $f(z)$ instead by looking at the Taylor series expansion of $\sin(z)$

$ f(z) = \frac{1}{(z)(z-\frac{z^3}{3!}+\frac{z^5}{5}+\cdots)}$ $= \frac{1}{(z^2)(1-\frac{z^2}{3!}+\frac{z^4}{5}+\cdots)}$ $ f(z) = \frac{1}{(z^2)}\left(1+\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)+\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)^2+\cdots\right)$ $ f(z) = \left(\frac{1}{z^2} + \frac{1}{z}\right) \left(1+\frac{z^2}{3!}+\cdot\right) = \frac{1}{z^2}+\frac{1}{z}+\cdots $

Rather than expanding at (what seems) an undefined point, a quick series alteration gives us a different "picture" of this function.

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    The last displayed identity for $f(z)$ is wrong.2012-05-21
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Multiply the function by z upstairs and downstairs:

$f(z)= \frac{z}{z^2⋅\sin z} = \frac{1}{z^2} \frac{ z }{ \sin z}$

Now, the second factor is a regular function with no poles. It remains just the factor in front which is:

$\frac{1}{z^2}$

A function with a single pole of order two, as expected.