Branch cut of the logarithm I have a function $F$ holomorphic on some open set, and I have $ F(0) = 1 $ and $ F $ is non-vanishing. I want to show that there is a holomorphic branch of $ \log(F(z)) $. 
Now, I'm getting confused. The principal branch of logarithm removes $ (-\infty, 0] $. But if the point 0 is missing from the plane, what happens when we take $ \log{F(0)} = \log{1} + 0 = 0 $? (I'm sure we can take the principal branch, because $ \exp(z) $ satisfies the conditions in question). 
Any help would be appreciated. Thanks
 A: If $F(0)=1$, then $\log F(x)$ may be defined uniquely and is holomorphic in any open connected and simply connected set $S$ containing the point $x=0$ as long as this open set avoids all points $x$ for which $F(x)=0$. After all, in one such an open set - the disk around the origin - the logarithm may be calculated as the Taylor expansion
$$\log F(x) = (F(x)-1) -\frac{(F(x)-1)^2}2 + \frac{(F(x)-1)^3}{3} - + \dots $$
The radius of the convergence is the minimum of the values $|x_0^{(i)}|$ where $x_0^{(i)}$ are all points such that $F(x_0^{(i)})=0$.
It doesn't matter that $\log F(0) = 0$. The logarithn is just equal to zero at the origin but because we're not calculating another logarithm of the logarithm, it doesn't hurt. Near the origin, $\log F(x)$ may be approximated by $F(x) -1$. The full series is above.
A: If the open set $U$ (where $F$ is defined) is not simply connected then $\log(F(z))$ may not exist (e.g. $F(z)=z$, $U=\mathbb{C}-0$). If $U$ is simply connected then $\log(F(z))$ exists by the Monodromy theorem: from a bit more general point of view, $\log$ is defined on a covering space $p:X\to\mathbb{C}-0$ ($X=\mathbb{C}$ and $p=\exp$), we have a map $F:U\to\mathbb{C}-0$, and since $U$ is $1$-connected, the map $F$ can be lifted to $X$.
A: That only means $F(z)$ cannot be 0, while $z$ can be 0.
