The Complex Logarithm of a Function

For an analytic function $f$ that does not vanish on a simply connected region, we may define its logarithm to be the function:

$$\log f=g(z):=\int_{y}\frac{f'}{f}dz+c_0.$$

Where $\gamma$ is some path starting at an arbitrary point in the region, and ending at $z$; while $c_0$ satisfies $e^{c_0}=f(z_0)$.

I believe that this logarithm should satisfy under certain conditions that: $$\log f=\log |f|+iarg(f).$$

Am I right, or this is too difficult in general?

Answer

The function $g$ satisfies $g' = \frac{f'}{f}$ in the given domain, so that
$$
(f e^{-g} )' = f' e^{-g} - f g' e^{-g} = 0 \\
\implies f e^{-g} = \text{const} = f(z_0) e^{-g(z_0)} = f(z_0) e^{-c_0} = 1 \, .
$$
Therefore $e^g = f$, i.e. $g$ is “a holomorphic logarithm” of $f$ in the domain. In particular
$$

f(z) = e^{g(z)} = e^{\operatorname{Re} g(z)} e^{ i \operatorname{Im}g(z)}
$$
which implies that
$$
|f(z)| = e^{\operatorname{Re}g(z)} \implies \operatorname{Re}g(z) = \log |f(z)|
$$
and that $ \operatorname{Im}g(z)$ is an argument of $f(z)$. So
$$
g(z) = \log |f(z)| + i \operatorname{arg}f(z)
$$

in the sense that $\operatorname{arg}f(z)$ is a continuous function which is an argument of $f(z)$ for each $z$.