Proving $f(z)=(e^{3iz}-3e^{iz}+2)/z^3$ has a simple pole at $0$ Let $f(z)=(e^{3iz}-3e^{iz}+2)/z^3$, this clearly has a singularity at $z=0$, how do you show that this is a simple pole? i.e. a pole of order one, every way it definity expands to something with a minimum power of $-3$, so I come to the conclusion there is a pole of order $3$?
And if this is the case is $f(z)+3/z$ holomorphic? Does this term cancel some part of the expansion I have missed that eliminates the pole?
 A: For $|z|\ll 1$, we have
$$
\begin{split}
e^{3iz} &= 1 + 3 i z - \frac{9}{2} z^2 + O(z^3) \\
e^{iz} &= 1 + i z - \frac{1}{2} z^2 + O(z^3)
\end{split}
$$
so 
$$ 
\begin{split}
\frac{1}{z^3} (e^{3iz} - 3 e^{iz} +2 )
&= \frac{1}{z^3} \left( 1 - 3 + 2 + (3 - 3) i z + (-\frac{9}{2} + \frac{3}{2}) z^2
+ O(z^3) \right) \\
&= - \frac{3}{z} + O(1)
\end{split}
$$
A: Consider $g(z) = e^{3iz} - 3e^{iz} + 2$. Then $g(0) = 1 - 3 + 2 = 0$, $g'(0) = 3i - 3i = 0$, and $g''(0) = (3i)^2 - 3i^2 = -9 + 3 = -6 \neq 0$. 
So $g(z)$ has a zero of order $2$ at $z = 0$, which means that $g(z)/z^3$ has a simple pole at $z = 0$.
Since $g(z) = g(0) + g'(0)z + {1 \over 2}g''(0)z^2 + ...$, the function $g(z)$ can be written as $-3z^2 + $ higher order terms. So ${\displaystyle f(z) = {g(z) \over z^3} = -{3 \over z}}$ plus an function that is analytic near $z = 0$. Hence ${\displaystyle f(z) + {3 \over z}}$ is analytic on a neighborhood of $z = 0$, and therefore on all of ${\mathbb C}$ since the function is explicitly defined in an analytic fashion except at $z = 0$.
