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 $n$?
On
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.