Residue Calculation I am stuck on what should be a trivial residue calculation. Any suggestions?
 Compute the residue of $\frac{e^{2iz}-1}{z^2}$ at $z=0$. Isn't it a double pole, i.e., shouldn't I be able to take the limit of the derivative of the function multiplied by $z^2$ to get $0$? ...for some reason the answer is $2i$?
 A: A function $f$ with a singularity at $z_0$ is said to have a pole of order $m$ at $z_0$ if, and only if, $\mathbb C\ni\lim \limits_{z\to z_0}((z-z_0)^mf(z))\neq 0$ and $\forall k\in \mathbb N(k< m \implies \lim \limits_{z\to z_0}((z-z_0)^kf(z)\not \in \mathbb C))$.
Since $\lim \limits_{z\to 0}\left(\dfrac{e^{2iz}-1}{z^2}\right)\not \in \mathbb C$ and $$\lim \limits_{z\to 0}\left((z-0)^1\dfrac{e^{2iz}-1}{z^2}\right)=\lim \limits_{z\to 0}\left(\dfrac{e^{2iz}-e^{2i\cdot 0}}{z-0}\right)=\left.\dfrac {\mathrm d}{\mathrm dz}\left(z\mapsto e^{2iz}\right)(z)\right|_{z=0}=\left.2ie^{2iz}\right|_{z=0}=2i\neq 0,$$ you have a pole of order $1$, not order $2$.
There is no need for power series or l'Hôpital.
But if you want to use power series to find the residue, then do it:
$$e^{2iz}-1=\sum \limits_{n=1}^\infty \left(\dfrac{(2i)^n}{n!}z^n\right).$$
Thus $$\dfrac{e^{2iz}-1}{z^2}=\sum \limits_{n=1}^\infty \left(\dfrac{(2i)^n}{n!}z^{n-2}\right).$$
Since the residue is the coefficient of $z^{-1}$ on this power series, (which comes for $n=1$), you get $\text{Res}(f,0)=2i$, as expected.
A: Use Taylor expansion for the exponential:
$$
e^{2iz}=\sum_{n=0}^{+\infty}\frac{(2iz)^n}{n!}
=1+2iz+z^2\overbrace{\sum_{n=2}^{+\infty}\frac{(2i)^nz^{n-2}}{n!}}^{g(z)}
$$
from which
$$
\frac{e^{2iz}-1}{z^2}=\frac{2iz+z^2g(z)}{z^2}=\frac{2i}{z}+g(z)\;,
$$
then you can conclude that $Res\left(\frac{e^{2iz}-1}{z^2},0\right)=2i$.
A: You will need to lhopital since the limit is indeterminant. i.e. $\frac{0}{0}$
