Question about branch of logarithm

  • #1
MathLearner123
16
3
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!
 
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  • #2
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.
 
  • #3
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.
 

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