order of the pole $0$ of function $f(z)=\frac{1}{z^{3/2}}$ 
Find order of the pole $0$ of $f(z)=\frac{1}{z^{3/2}}$.

I think it is $2$ because 
\begin{align*}
z^{3/2}\big|_{0}&=0\\
(z^{3/2})'\big|_{0}&=0\\
(z^{3/2})''\big|_{0}&\neq 0,
\end{align*}
but I'm not sure. Could someone explain to me, please?

As Chappers pointed out this function is not analytic on any annalus centred at $z=0$. It's a multivalued function defined by:
\begin{align*}
 z^{-\frac32} = e^{Ln(z^{-\frac32})} = e^{-\frac 32 (\ln|z| +i\arg(z)+2p\pi i)} = |z|^{-\frac32} e^{\frac{-3i}{2}(\arg(z)+2p\pi)},
\end{align*}
where $p \in \mathbb Z$. 
Another way to see that something odd is going on is to evaluate $f$ along the circle given by $e^{i \theta}$. At $\theta =0$ we have that $f(e^{i0})=e^0=1$ (naivly). But at $\theta = 2\pi$ (the same point in th complex plane) we get $f(e^{i2\pi}) = e^{-i3\pi} = -1$ (which is the main concern here).
