# Math Help - Residue problem

1. ## Residue problem

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.

2. Originally Posted by boredaxel
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 $\frac{1}{\sin z}$ (from which it immediately follows that the residue is equal to 1/6).

3. 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?

4. Originally Posted by boredaxel
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 $\frac{1}{\sin z}$ has a pole of order 1 at z = 0 since $\lim_{z \to 0} \frac{z}{\sin z}$ is finite (and is equal to 1). Furthermore, $\text{Res}\left( \frac{1}{\sin z}, z = 0\right) = 1$.

Therefore $\frac{1}{\sin z} = \frac{1}{z} + a + bz + cz^2 + ....$

Therefore $\text{Res}\left( \frac{1}{z^2 \sin z}, z = 0\right) = b$.

To get the value of b, substitute $\sin z = z - \frac{z^3}{6} + \frac{z^5}{120} - ....$ into $\frac{1}{\sin z} = \frac{1}{z} + a + bz + cz^2 + ....$ and re-arrange:

$1 = \left(z - \frac{z^3}{6} + \frac{z^5}{120} - .... \right) \left(\frac{1}{z} + a + bz + cz^2 + ....\right)$

Expand and equate coefficients:

Coefficient of z: a = 0.

Coefficient of $z^2$: $0 = b - \frac{1}{6} \Rightarrow b = \frac{1}{6}$.

5. What can't Wolfram Alpha do?

6. Originally Posted by chiph588@
What can't Wolfram Alpha do?
What can't you do? - Wolfram|Alpha