Find order of pole $\frac{e^z -1}{z^2 +4}$, about $z=2i$.

Question

$$\frac{e^z -1}{z^2 +4},\quad\text{about $z=2i$.}$$ The textbook I'm reading isn't specific about these case, only gives basic examples. Basically to find pole I'd have to expand a Laurent series about some point, in this case $z=2i$, I did $$ \frac{e^{2i} e^{z-2i} -1}{(z-2i)^2 +4-(4z-4)}$$ and from there just expanded $e^{z-2i}$ as a series, just see which powers (without caring about constants) of $z-2i$ was left after canceling, giving me a pole of order 2. Answer however says of order 1.

Answer 1

$z^2+4$ has in $2i$ a simple zero and $e^{2i}-1 \ne 0$, hence $2i$ is a simple pole of $\displaystyle\frac{e^z -1}{z^2 +4}$.