Branch of the n-th root of a holomorphic function is holomorphic It is true that given a domain $D$ and a function $f$ that is holomorphic and not vanishing in $D$, then a branch $\varphi$ (with $\varphi$ continuous) of $f^{1/n}$ in $D$ is holomorphic in $D$ (here is a proof for that statement ). 
My question is, does the statement remain true if we remove the non-vanishing condition? I haven't been able to find a counterexample or a proof on that more general case.
 A: If $f$ is holomorphic in $D$ and $\varphi$  continuous in $D$ with $\varphi^n = f$ then $\varphi$ is also holomorphic in $D$. Under these conditions it is not necessary to assume that $f$ does not vanish in $D$.
First consider $\varphi$ restricted to $D' =  \{ z \in D : f(z) \ne 0 \}$: $\varphi|_{D'}$ is holomorphic because it is locally equal to $\exp(\frac 1n \log z)$ for some branch of the logarithm. Or more elementary: If $f(z_0) \ne 0$ then for $z$ close to $z_0$:
$$
 \frac{f(z)-f(z_0)}{z-z_0)} = \frac{\varphi(z)^n-\varphi(z_0)^n}{z-z_0}
 \\
= \frac{\varphi(z)-\varphi(z_0)}{z-z_0} \cdot \left(
\varphi(z)^{n-1} + \varphi(z)^{n-2}\varphi(z_0) + \ldots \varphi(z)\varphi(z_0)^{n-2} + \varphi(z_0)^{n-1}
\right)
$$
And since the term in parentheses on the right has the non-zero limit
$(n-1) \varphi(z_0)^{n-1} $ for $z\to z_0$ it follows that the limit
$$
\lim_{z\to z_0} \frac{\varphi(z)-\varphi(z_0)}{z-z_0}
= \frac{1}{(n-1) \varphi(z_0)^{n-1}} f'(z_0)
$$
exists, i.e. $\varphi $ is differentiable at $z_0$.
It remains to consider the points in $D$ with $f(z_0)=0$. At all these points, $\varphi|_{D'}$ has an isolated singularly, and that is removable because $\lim_{z\to z_0} \varphi(z) = 0$, according to Riemann's theorem on removable singularities.
It follows that $\varphi|_{D'}$ can be holomorphically extended to $D$, and that extension does of course coincide with $\varphi$ (since both functions are continuous).
This proves that $\varphi$ is in holomorphic in $D$.
A: No, it doesn't. Take $f(z)=z$ and $n=2$, for instance. There is no holomorphic functions (or even a continuous one) from $\mathbb C$ into $\mathbb C$ such that $\varphi^2=f$. If there was, and if $k$ was the order of $0$ as a zero of $\varphi$, then $0$ would be a zero of otder $2k$ of $f$. But $0$ is a zero of order $1$ of $f$.
