Is the principal root defined at the right half-plane continuous at $0$?

Question

I'm reading a proof of a theorem. In the proof the author says that the principal root $z^{1- \delta/2}$ (that is, the function $z^{1- \delta/2}:=\exp(\mbox{Log}(z)(1- \delta/2))$, where $\mbox{Log}$ is the principal branch of logarithm function) for $ \delta \in (0,1]$ is continuous at the boundary of the right half-plane. But I think that maybe there is problem at $0$. Is the principal root $z^{1- \delta/2}$ continuous at $0$? Can we say that $\exp(z^{1- \delta/2})$ is continuous at $0$?

Answer 1

Let $z = re^{i\theta}$ and let $\epsilon = 1 - \frac \delta 2$. Note that $\frac 12 \le \epsilon < 1$. Now $$|\exp(\operatorname{Log}(z)\epsilon)| = r^\epsilon$$

As $z \to 0, r = |z| \to 0$, and therefore $r^\epsilon \to 0$. That is, $$\exp(\operatorname{Log}(z)\epsilon) \to 0$$

While $0$ is not in the domain of $\operatorname{Log}$, and therefore the full expression, this is a straight-forward continuous extension.