Question about branch of logarithm

[MathLearner123]

I have a question about Daniel Fischer's answer here

Why the function $g(w)$ is well-defined on $\mathbb{D} \setminus \{0\}$? I don't understand how $\log$ function works here and how a branch of $\log$ function can be defined on whole $\mathbb{D} \setminus \{0\}$. For example principal branch of logarithm is defined on $\mathbb{C} \setminus \mathbb{R}_{-}$ so can not be used for $\mathbb{D} \setminus \{0\}$. Thanks!

 

[fresh_42]

For all who do not want to switch between pages, here is the quotation (link):

Consider the function
$$h(z) = f(z)\cdot e^{-2\pi iaz}$$
on the upper half-plane. We have
$$h(z+1) = f(z+1)\cdot e^{-2\pi i a (z+1)} = \bigl(f(z)e^{2\pi i a}\bigr)e^{-2\pi ia}e^{-2\pi i az} = f(z)e^{-2\pi i az} = h(z),$$
i.e. $h$ is periodic with period $1$. Now define $g \colon \mathbb{D}\setminus \{0\} \to \mathbb{C}$ by
$$g(w) := h\left(\frac{\log w}{2\pi i}\right).$$
By the periodicity of $h$, the value of $g(w)$ is independent of the choice of the branch of the logarithm, hence $g$ is well-defined. In a small enough neighbourhood of any $w \in \mathbb{D}\setminus\{0\}$, there is a holomorphic branch of the logarithm, hence $g$ is holomorphic.
By construction, we have
$$h(z) = g\left(e^{2\pi i z}\right)$$
and hence
$$f(z) = e^{2\pi i az}g\left(e^{2\pi iz}\right),$$
as desired.

 

[WWGD]

I understood Fischer was referring to its existence, of the log, in a neighborhood of $D-\{0\}$. Just take an open ball that avoids the origin.