Algebraic topology - angle function

162 Views Asked by At

I just know "if $f$ is continuous, $f: [a, b]\rightarrow S^1$, then there is a continuous function $g: [a, b] \rightarrow \mathbb R$, such that $f (x) = e ^ {g (x)}$, for all $x$ in $[a, b]$"

I want to solve the following exercise:

If $f: U \rightarrow \mathbb C \smallsetminus\{0\}$ is continuous, $U$ open subsset of $\mathbb C$. Prove that there exists a continuous function $g: U \rightarrow \mathbb C$ such that $f (z) = e ^ {g (z)}$, for all $z \in U$ if, and only if, for every closed path $c$ in $U$ has $ind(f \circ c, 0) = 0$.

Is it possible? how?

3

There are 3 best solutions below

2
On

Here's a hint: Think about the topological characterization of domains $U \subset \mathbb{C}$ on which a logarithm can be defined. The function $g(z)$ here is a logarithm of $f(z)$, and hence a related topological description will emerge.

1
On

The logarithm of a function g is defined as the integral $$\int_{\gamma} \frac {dg(z)}{z} $$

Where $\gamma$ is a curve in the domain.

This integral is well-defined, i.e., independent of path only in simply-connected regions that do not wind around the origin.

0
On