# Extending the cube root function to $\mathbb{C}$

On $\mathbb{R}$, the cube-root function (call it $f(x)$) is a well-defined single-valued function which is $C^\infty$ except at the origin. ($f(27)=3$, $f(-27)=-3$, etc.)

The complex cube root (call it $u(z)$) is triple-valued with branch points at $0$ and $\infty$, so any continuous single-valued branch requires us to exclude some curve connecting $0$ and $\infty$ from the domain. The origin-centered circle $C_R$ of radius $R$ is thus mapped by $u$ (minus a point excluded from the domain) to an arc subtending an angle of $2\pi/3$ at the origin.

It seems to me on geometrical grounds that this means no choice of domain for $u$ can make it restrict to $f$ on the real axis. The image of $C_R\cap \mathbb{R}$ under $f$ includes both $\sqrt[3]{R}$ and $-\sqrt[3]{R}$, but $u$ maps the whole of $C_R$, minus the excluded point, to an arc subtending an angle of less than $\pi$, which thus can't contain both a real number and its negative.

Do you agree? Is it safe to say that the real cube root function (even excluding the origin) has no continuous analytic extension to (any connected region in) $\mathbb{C}$? If not, how should this statement be qualified?

• The cubic root function is just $e^{log{1/3z}}$, it is better to think this not in terms of branch cuts but as a function in the Riemann Surface. – Kerry Oct 11 '11 at 19:30
• @Changwei Zhou: Did you mean $e^{1/3 \log z}$? – Rasmus Oct 11 '11 at 19:51
• @Rasmus, yes. But I cannot edit my own comment. – Kerry Oct 11 '11 at 23:40

Sure. If you start with $f$ restricted to the positive axis, and continue it analytically along a curve in the upper half plane to $-1$, you will obtain $e^{i\pi/3}$ (not $-1$) as the cube root of $-1$.