Difficulties performing Laurent Series expansions to determine Residues The following problems are from Brown and Churchill's Complex Variables, 8ed. 
From §71 concerning Residues and Poles, problem #1d:
Determine the residue at $z = 0$ of the function $$\frac{\cot(z)}{z^4} $$
I really don't know where to start with this. I had previously tried expanding the series using the composition of the series expansions of $\cos(z)$ and $\sin(z)$ but didn't really achieve any favorable outcomes. If anyone has an idea on how I might go about solving this please let me know. For the sake of completion, the book lists the solution as $-1/45$.
From the same section, problem #1e
Determine the residue at $z = 0$ of the function $$\frac{\sinh(z)}{z^4(1-z^2)} $$
Recognizing the following expressions: 
$$\sinh(z) = \sum_{n=0}^{\infty} \frac{z^{(2n+1)}}{(2n +1)!}$$
$$\frac{1}{1-z^2} = \sum_{n=0}^{\infty} (z^2)^n $$
I have expanded the series thusly:
$$\begin{aligned} \frac{\sinh(z)}{z^4(1-z^2)}  &= \frac{1}{z^4} \bigg(\sum_{n=0}^{\infty} \frac{z^{(2n+1)}}{(2n +1)!}\bigg) \bigg(\sum_{n=0}^{\infty} (z^2)^n\bigg) \\ &=  \bigg(\sum_{n=0}^{\infty} \frac{z^{2n - 3}}{(2n +1)!}\bigg) \bigg(\sum_{n=0}^{\infty} z^{2n-4} \bigg) \\\end{aligned}  $$
I don't really know where to go from here. Any help would be great.
Thanks.
 A: A related problem. Lets consider your first problem
$$ \frac{\cot(z)}{z^4}=\frac{\cos(z)}{z^4\sin(z)}. $$
First, determine the order of the pole of the function at the point $z=0$, which, in this case, is of order $5$. Once the order of the pole has been determined, we can use the formula
$$r = \frac{1}{4!} \lim_{z\to 0} \frac{d^4}{dz^4}\left( z^5\frac{\cos(z)}{z^4\sin(z)} \right)=-\frac{1}{45}. $$
Note that, the general formula for computing the residue of $f(z)$ at a point $z=z_0$ with a pole order $n$ is 

$$r = \frac{1}{(n-1)!} \lim_{z\to z_0} \frac{d^{n-1}}{dz^{n-1}}\left( (z-z_0)^n f(z) \right) $$

Note: If $z=z_0$ is a pole of order one of $f(z)$, then the residue is
$$ r = \lim_{z\to z_0}(z-z_0)f(z). $$
A: $$\cos z=1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\;,\;\;\;\sin z=z-\frac{z^3}{6}+\frac{z^5}{120}-\ldots\implies$$
$$\cot z=\frac{\left(1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\right)}{z\left(1-\left(\frac{z^2}{6}-\frac{z^4}{120}\right)+\mathcal O(z^6)\right)}=$$
$$=\frac{1}{z}\left(1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\right)\left(1+\left(\frac{z^2}{6}-\frac{z^4}{120}\right)+\left(\frac{z^2}{6}-\frac{z^4}{120}\right)^2+\ldots\right)=$$
$$=\frac{1}{z}\left(1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\right)\left(1+\frac{z^2}{6}-\frac{z^4}{120}+\frac{z^4}{36}-\frac{z^6}{360}+\frac{z^8}{120^2}+\ldots\right)$$
$$=\frac{1}{z}\left(1-\frac{z^2}{3}-\frac{z^4}{45}+\ldots\right)=\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}+\mathcal O(z^5)$$
Multipying the above by $\,\displaystyle{\frac{1}{z^4}}\;$ renders the residue $\,\displaystyle{-\frac{1}{45}}\,$ .
Notice we only use the powers necessary to calculate the coefficient of $\,z^{-1}\,$ . All the rest is unimportant to us.
