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.
Best Answer
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.