Classification of isolated singularity I have found the Laurent Series of $\displaystyle \frac{\sin3z}{z^5}$  about $z=0$ to be: 

$$ \sum_{n=-2}^{\infty} \frac{(-1)^n3^{2n+5}z^{2n}}{(2n+5)!}$$

I now have to show that $\displaystyle \frac{\sin3z}{z^5}$ has a pole of order $4$ at $z=0.
$

How do I go about classifying this seeing as $z$ is to the power of $2n$ as opposed to the usual $n$?
We don't have to use the series expansion to see that $\sin(3z)/z^5$ has a pole of order four. From Calculus, we learned that 
$$
\lim_{z\to 0}\frac{\sin(z)}{z} = 1
$$
and diverges to infinity when the degree of the denominator is positive and odd and DNE for positive even degrees.
For a residue at $z=0$ for $\sin(3z)/z^5$, we have
$$
\lim_{z\to 0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\frac{\sin(3z)}{z^5}(z-z_j)^n =
\lim_{z\to 0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\frac{\sin(3z)}{z^5}z^n
$$
where $z_j = 0$. It should be pretty apparent now what order the pole at $z=0$ is.
If you are interested in poles at $z=0$, you only need the terms in the series with a negative exponent of $z$.
Write down the first few terms and you'll see what happens.

Also, you can define
$$
a_n
=
\frac{(-1)^{n/2}3^{n+5}}{(n+5)!}
$$
for even $n$ and $a_n=0$ for odd $n$.
Then your series is $\sum_{n=-4}^\infty a_nz^n$, if that form is what you are after.