Analytical continuation of $z^\alpha$

Question

As $x^{\alpha}, \alpha\in\mathbb{R}$ is analytical in $(0,\infty)$, can I find an analytical continuation of $x^{\alpha}$, say $z^{\alpha}$ in $\mathbb{C}\backslash (-\infty,0]$?

Answer 1

Sure, $z^\alpha = \exp(\alpha \log(z))$ where one uses the analytic branch of $\log(z)$ on $\mathbb{C} \setminus (-\infty,0]$ that agrees with $\ln(x)$ on $(0,\infty)$.