Problem
Define g to be the function $g(z)=re^{\frac{i\theta}{2}}$ if $z=re^{i\theta}$ with $r>0$ and $-\pi<\theta\le\pi$, and $g(z)=0$ when $z=0$.
Is $g$ continuous from $\mathbb{C}\longrightarrow\mathbb{C}$?
Progress
Claim: $g$ is not continuous from $\mathbb{C}\longrightarrow\mathbb{C}$.
[It seems that all the function seems to be doing is halving $arg(z)$ for each $z\in\mathbb{C}$.]
If we consider points on the negative real line, i.e. $z=-a$ for $a\in\mathbb{R}$, then let $\epsilon=\frac{a}{2}$. If $g$ is continuous, then $\exists \delta>0$ such that $g(B_{\delta}(z))\subset B_{\epsilon}((g)z)$. Clearly $g(-a-(i\delta/2))\in g(B_{\delta}(-a))$ does not lie in $B_{a/2}(g(-a))$ for any $\delta>0$ as $-a$ is mapped to $ai$ and $-a-(i\delta/2)$ is mapped to a point in the lower half-plane.
We can conclude then that $g$ is not continuous from $\mathbb{C}\longrightarrow\mathbb{C}$.
Thoughts
Is this proof valid (if I play around to make it a little more formal), i.e. is the reasoning behind the claim correct?
Any verification/objection would be appreciated. Thanks, TJO.