If $z=re^{i\theta}$, then $\sqrt{z}=\sqrt{r}e^{i\frac{\theta}{2}}$.
Isn't this a well defined function on the whole complex plane?
Why do we need to define this as the function $\mathbb C \setminus \{x \in \mathbb R \mid x \le 0\} \to \{re^{i\theta} \mid -\pi < \theta < \pi\}$?
It's "well-defined" if you, for every $z$, pick some $\theta_z$ such that $z=|z|e^{i\theta_z}$, but the function above won't be continuous.
Let $x\in (-\infty,0)$ and consider $x+i\varepsilon$ and $x-i\varepsilon$. The square-root you seem to define of the first thing tends to $i\sqrt{|x|}$ as $\varepsilon\to 0,$ while the second one tends to $-i\sqrt{|x|}$.