I have a Big-O Problem I want to show: $(1-3z)^{3/2}$ is O(1-3z) as $z\rightarrow 1/3$ where $z \in \mathbb{C}$
I would like to be able to write: $\displaystyle \frac{(1-3z)^{3/2}}{1-3z}=(1-3z)^{1/2}$, and then show that as $z \rightarrow \frac{1}{3}, (1-3z)^{1/2} \rightarrow 0$
But because there is a singularity at $\frac{1}{3}$, it is not clear how one could take a limit as $z \rightarrow \frac{1}{3}$

You are probably expected to write something like, when $z$ is in $\mathbb C\setminus\{\frac13\}$,
$$
\left|\frac{(1-3z)^{3/2}}{1-3z}\right|=|1-3z|^{1/2}\stackrel{z\to1/3}{\longrightarrow}0.
$$
Unfortunately for this "proof", no continuous function $z\mapsto(1-3z)^{1/2}$ can exist in a neighborhood of $z=\frac13$, likewise for $z\mapsto(1-3z)^{3/2}$.
