Given that $\frac{z^a}{z^b}$ for $b\in\mathbb{N}$ and $a\in\mathbb{R}^{+} \setminus \mathbb{N}$ has a pole of order $b$ at the origin, I chose to employ the derivative-limit formula.
$$\text{Res}_{z=0} \frac{z^a}{z^b} = \frac{1}{(b-1)!} lim_{z \to 0} \frac{d^{b-1}}{dz^{b-1}} \left(z^b \frac{z^a}{z^b}\right)$$
I then assert that calculation of the residue is no more than evaluating this limit. $$\frac{1}{(b-1)!} \text{lim}_{z \to 0} \frac{d^{b-1}}{dz^{b-1}} \left(z^a\right) = \frac{(a)(a-1)...(a+2-b)}{(b-1)!} \text{lim}_{z \to 0} z^{a+1-b} = \begin{cases} 0 & \text{if} & a+1 > b \\ \infty & \text{if} & b > a+1 \\ \frac{(a)(a-1)...(a+2-b)}{(b-1)!} & \text{if} & b=a+1 & \text{(impossible given that $a \in \mathbb{R}^+ \setminus \mathbb{N}$)} \end{cases}$$
When I had originally studied complex analysis, I had never encountered an infinite residue, so I believe that I have made a rather obvious mistake somewhere in my reasoning. If a residue were infinite, would that imply that the Laurent series for the function would not exist or diverge everywhere in an annular region around the origin?