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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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Basically, computers are stupid. – anon Apr 18 '12 at 0:09
so I am right? The residue at $z = 0$ is $0$, the residue at $z = 1$ is $\frac{-1}{\pi}$ and so on? – Tyler Hilton Apr 18 '12 at 0:12… – Aryabhata Apr 18 '12 at 0:14
Hi Aryabhata. I added that into my post. I saw that too. This is exactly my question. We know other poles and residues exist so why is that wolfram just identifies 0 as a pole. I mean if it can identify $n\pi$ as poles for $\frac{1}{\sin{z}}$ then it should be able to do that if you multiply the denominator by z. – Tyler Hilton Apr 18 '12 at 0:19
FWIW, Maple 16 says: > singular(1/(zsin(z))); {z = 0}, {z = Pi_Z1} – Robert Israel Apr 18 '12 at 1:14
up vote 0 down vote accepted

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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What am I looking at though? This is just expansion around z=0 no? – Tyler Hilton Apr 18 '12 at 4:11
You are correct on everything you've said except that zero is not the only pole, n* pi for n integer (and not zero) gives a pole of order 1, you can verify this (as you have) with a Laurent Series expansion or by famous results/theorems on poles/singularities. – Arbias Hashani Apr 19 '12 at 13:35
The last displayed identity for $f(z)$ is wrong. – Did May 21 '12 at 11:54

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:


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

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