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

$$\frac{e^z -1}{z^2 +4},\quad\text{about z=2i.}$$ 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.

• $$\frac{e^z-1}{z^2+4}=\frac{\frac{e^z-1}{z+2i}}{z-2i}=\frac{h(z)}{z-2i}$$ with $h(2i) \neq 0$. Commented Jan 22, 2019 at 6:37

$$z^2+4$$ has in $$2i$$ a simple zero and $$e^{2i}-1 \ne 0$$, hence $$2i$$ is a simple pole of $$\displaystyle\frac{e^z -1}{z^2 +4}$$.
• Thank you again Fred. If the numerator was $= 0$, then what would we have to do? Say $\frac{z^3-1}{(z-1)^3}$ at z = 1 Commented Jan 22, 2019 at 6:44
• nevermind, I understand why you cared about the numerator. in this case, it would e $\frac{0}{0^3}$ and have one of the pole order is removable singularity, so pole would be order of 2. Commented Jan 22, 2019 at 7:08
• $\frac{z^3-1}{(z-1)^3}=\frac{z^2+z+1}{(z-1)^2}$, hence $\frac{z^3-1}{(z-1)^3}$ has in $1$ a pole of order $2$.