Holomorphic branch of the $n^{\mathrm{th}}$ root of $f$ The problem states to prove that if $h$ is a branch of $f^{1/n}$ for integer $n > 0$ (i.e. $h(z)^n = f(z)$ for $z \in G$, $h$ continuous), then $h$ is holomorphic, where $f$ is a holomorphic function on an open connected subset $G$ of $\mathbb{C}$ and $f \neq 0$ on $G$.
I'm not sure where to start; the Cauchy-Riemann equations seem to be a bad route. I was thinking of trying to prove by contradiction that if $h$ is not complex differentiable at a point, then $f$ must have a point where it's not complex differentiable, but that hasn't been fruitful. Algebraic manipulation of the limit definition of a derivative doesn't seem to be a good idea either.
Any tips?
 A: Let $a\in G$. We need to show that $h$ is differentiable at $a$.
As $G$ is open there exists a disc $D(a,r)\subset G$. Now as $f$ does not vanish in $G$, 
and hence in $D(a,r)$, we define the function
$$
F(z)=\int_{[a,z]}\frac{f'}{f}.
$$
Then $F$ is holomorphic in $D(a,r)$ and the product $\exp(-F(z))f(z)$ 
is constant in $D(a,r)$, as
$$
\big(\exp(-F(z))f(z)\big)'=-\exp(-F(z))F'(z)f(z)+\exp(-F(z))f'(z)=0.
$$
Let $c=\exp(-F(z))f(z)$ or equivalently $f(z)=c\exp(F(z))$. Clearly $c\ne 0$, as $f$ does not vanish.
Set now 
$$
g(z)=c_1\exp\left(\frac{F(z)}{n}\right).
$$ 
Then $g$ is holomorphic in $D(a,r)$ and choose $c_1$ so that $\left(g(z)\right)^n=f(z)$ - i.e., $c_1^n=c$.
But $\big(h(z)\big)^n=f(z)$, as well, and therefore
$$
\left(\frac{h}{g}\right)^n=1, \quad z\in D(a,r).
$$
Thus, for all $z\in D(a,r)$:
$$
\frac{h(z)}{g(z)}\in\{\mathrm{e}^{2k\pi i/n}: k=0,\ldots,n-1\},
$$
and as the latter set is discrete and $h/g$ is continuous on the connected set $D(a,r)$, then $f/g$ is constant, and hence $h$ is holomorphic in $D(a,r)$. But $a$ was chosen arbitrarily, and hence $h$ is holomorphic on the whole $G$.
A: Hint:
try with branch of logarithmic function.
