Laurent Series Expansion computing terms

I need to compute the -5th term to the 5th term of the Laurent expansion of $(\cos(z))^2/\sin(z)$. I know that I can make this into $\csc(z)-\sin(z)$ but I wouldn't know what to do with the $\csc(z)$ in terms of a series. Also, I feel like several of these Laurent expansion coefficients would be zero. Could someone give me some help with this one I am struggling a little bit with Laurent series in general. I also need to find the domain of convergence.

$$\cos^2z = 1-z^2+\frac{z^4}{3}-\frac{2z^6}{45}+\cdots$$

The function $\frac{1}{\sin z}$ has a simple pole at $z=0$ and it is analytic in a punctured disc around $z=0$, so it's Laurent series is $\frac{a_{-1}}{z}+a_0+a_1z + \cdots$ and it converges to $\frac{1}{\sin z}$ in a punctured disc around $z=0$.
Since $\sin z = z-\frac{z^3}{6}+\frac{z^5}{120} + \cdots$ we have have for $z\ne0$: $$1=\Big( \frac{a_{-1}}{z}+a_0+a_1z + \cdots \Big)\Big( z-\frac{z^3}{6}+\frac{z^5}{120} -\frac{z^7}{5040}+ \cdots \Big)$$ or: $$1=\Big( a_{-1}+a_0z+a_1z^2 + \cdots \Big)\Big( 1-\frac{z^2}{6}+\frac{z^4}{120} -\frac{z^6}{5040}+ \cdots \Big)$$

By comparing the coefficients of the powers of $z$ in both sides you get: $$a_{-1}=1,\space a_0=0, \space a_1=\frac{1}{6}, \space a_2=0, \space a_3=\frac{7}{360}, \space a_4=0, \space a_5=\frac{31}{15120}$$
And so we have: $$\frac{1}{\sin z} = \frac{1}{z} + \frac{z}{6} + \frac{7z^3}{360} + \frac{31z^5}{15120} + \cdots$$
And therefore: $$\frac{\cos^2z}{\sin z} = \Big( \frac{1}{z} + \frac{z}{6} + \frac{7z^3}{360} + \frac{31z^5}{15120} + \cdots \Big) \Big( 1-z^2+\frac{z^4}{3}-\frac{2z^6}{45}+\cdots \Big)$$

$$= \frac{1}{z}-\frac{5z}{6} + \frac{67z^3}{360}-\frac{19z^5}{3024} + \cdots$$

• Why wouldn't z=pi be a pole? Also, is there any way to find an exact radius for the disk it converges in? – user153009 Mar 22 '15 at 12:49
• $z=\pi$ is a pole. In fact $z=\pi k$ are poles. The radius of convergence is $\pi$ – benji Mar 22 '15 at 13:29
• Okay, thank you very much. And lastly, poles always occur when the denominator is zero right? I know the definition of a pole but is this always the case? – user153009 Mar 22 '15 at 13:40
• No, consider for example $\frac{z}{\sin z}$ where the denominator is $0$ at $z=0$ but the singularity is removable, or $\frac{1}{\sin \frac{1}{z}}$ where the denominator inside the $\sin$ is $0$ at $z=0$ but the singularity is essential – benji Mar 23 '15 at 1:07

An idea:

$$\frac{\cos^2x}{\sin x}=\frac{\left(1-\frac{x^2}2+\frac{x^4}{24}-\ldots\right)^2}{x-\frac{x^3}6+\frac{x^5}{120}-\ldots}=\frac{1-x^2+\frac{x^4}3+\mathcal (x^6)}{x\left(1-\frac{x^2}6+\mathcal O(x^4)\right)}=$$

$$=\frac1x\left(1-x^2+\frac{x^4}3+\mathcal O (x^6)\right)\left(1+\frac{x^2}6+\frac{x^4}{36}+\mathcal O(x^6)\right)$$