a) If $D \subset \mathbb C\setminus\{0\}$ is an open connection. Show that if $\theta_1$ and $\theta_2$ are argument branches in $D$, then there is $k \in Z$ such that $\theta_1(z) = \theta_2(z) + 2k\pi,\, \forall z \in D$.
b) Show that there is no logarithm branch in $\mathbb C \setminus\{0\}$.
I don't think I understand this matter.
a) What I know: An argument branch in $D \subseteq \mathbb C - \{0\}$ is a continuous function $f:D \to \mathbb R$ such that \begin{align} z &= |z| \, \bigl (\cos(f(z)) + i \sin(f(z)) \bigr) \\ \implies z &= |z| \, \bigl( \cos(\operatorname{Im}(L(z)) ) + i \sin(\operatorname{Im} (L(z))) \bigr) \end{align} but $\theta = \operatorname{Im} (L(z))$, and $z = e^{L(z)} = |z| \, \bigl( \cos(\theta(z)) + i \sin(\theta(z)) \bigr)$.
I also know that: Let $D \subseteq \mathbb C\setminus \{0\}$ be an open one are equivalent:
- There is a branch $L$ of $\log z$ in $D$.
- There is a branch $\theta$ in $\arg z$ in $D$.
And by hypothesis there are $\theta_1$ and $\theta_2$, so there are \begin{align} L(z_1) = \log(z_1) + i\theta(z_1) &\implies e^{L(z_1)} = e^{\log(z_1)} e^{ i\theta(z_1)}, \\ L(z_2) = \log(z_2) + i\theta(z_2) &\implies e^{L(z_2)} = e^{\log(z_2)} e^{i\theta( z_2)}. \end{align} That's it, I completely crashed.
Is question b) correct? Not the opposite of the proposition Let $f: D \subseteq \mathbb C\setminus \{ 0 \} \to \mathbb C $, open connected $D$, a logarithm branch on $D$. So every logarithm branch in $D$ is of the form $f(z) + 2k\pi i, \forall \in \mathbb Z$.
Grateful for the attention.