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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

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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.

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Compute a function $G(w)$ your branch of $\log F(w)$ ... Another way to do this is to let $G(w)$ be the integral of $F'(z)/F(z)\,dz$ along a path from $0$ to $w$ in that (open simply connected) region. The answer is independent of the choice of path. – GEdgar May 18 '11 at 13:07
Thanks for your comment, but when you mention an "answer", what is the question you're trying to answer? – Luboš Motl May 18 '11 at 13:37

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 $X$.

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That only means $F(z)$ cannot be 0, while $z$ can be 0.

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