How do I find the residue of 1/(z^2 sinz) ? It seems to me that it has a 3rd order singularity at z=0, but I cant find the residue. Thanks for any help rendered.
It has a pole of order 3 at z = 0. And there is a standard formula (it will be in your textbook or class notes) for finding the residue at a pole ....
Alternatively, you could find and use the first few terms of the Laurent series of $\displaystyle \frac{1}{\sin z}$ (from which it immediately follows that the residue is equal to 1/6).
I tried using the formula but I got this equation eventually (attached along this reply). I could not eliminate the sinz at the denominatior, causing my ans to tend to inf as z tends to 0. Have I used the formula wrongly or is there another formula to get the residue?
Following up from my earlier reply:
Note that $\displaystyle \frac{1}{\sin z}$ has a pole of order 1 at z = 0 since $\displaystyle \lim_{z \to 0} \frac{z}{\sin z}$ is finite (and is equal to 1). Furthermore, $\displaystyle \text{Res}\left( \frac{1}{\sin z}, z = 0\right) = 1$.
Therefore $\displaystyle \frac{1}{\sin z} = \frac{1}{z} + a + bz + cz^2 + .... $
Therefore $\displaystyle \text{Res}\left( \frac{1}{z^2 \sin z}, z = 0\right) = b$.
To get the value of b, substitute $\displaystyle \sin z = z - \frac{z^3}{6} + \frac{z^5}{120} - .... $ into $\displaystyle \frac{1}{\sin z} = \frac{1}{z} + a + bz + cz^2 + .... $ and re-arrange:
$\displaystyle 1 = \left(z - \frac{z^3}{6} + \frac{z^5}{120} - .... \right) \left(\frac{1}{z} + a + bz + cz^2 + ....\right) $
Expand and equate coefficients:
Coefficient of z: a = 0.
Coefficient of $\displaystyle z^2$: $\displaystyle 0 = b - \frac{1}{6} \Rightarrow b = \frac{1}{6}$.
What can't Wolfram Alpha do?