Question

Let Ω be a subset of the complex plane (C) that is open, and f : Ω → C is a holomorphic function, and z⁰ belongs to Ω. Suppose f(z⁰) ≠ 0.

a) Show that there is an open neighborhood U within Ω of z⁰, and g : U → C is a holomorphic function such that e^g(z) = f(z) for z in U.
b) What is the question related to this situation?

Answer 1

The student's question is about finding a holomorphic function g on an open neighborhood U such that e^g(z) equals the given holomorphic function f(z) at a point z where f(z) is not equal to zero, indicating the local existence of a complex logarithm for f.

Step-by-step explanation:

The question is asking to prove that if f is a holomorphic function on an open subset Ω of the complex plane, and there exists a point z⁰ in Ω such that f(z⁰) ≠ 0, then we can find an open neighborhood U around z⁰ such that there is another holomorphic function g on U for which e^g(z) = f(z) for all z in U. The question is related to the theory of holomorphic functions, also known as analytic functions, which are complex functions that are differentiable at all points in their domain.

To answer part a), we use the fact that holomorphic functions possess a power series expansion around any point in their domain. Since f(z⁰) ≠ 0, we can consider the power series of f around z⁰. By the Existence of Logarithm theorem for holomorphic functions, there exists a local branch of the logarithm of f, and hence we can define g(z) as the logarithm of f(z) in this neighborhood. Since e^g(z) and f(z) both have power series representations, it follows from the uniqueness of power series that e^g(z) = f(z) for all z in the neighborhood U.