Currently, I am trying to solve the following exercise involving the definition of branches of logarithms:
Def: Let $D \subseteq \mathbb C$ be a domain, we call every continuous function $L : D \to \Bbb C$ a branch of the logarithm if it satisfies $$ e^{L(z)} = z \quad \forall z \in G. $$
Exercise: Let $D \subseteq \mathbb C$ be a simply connected domain not containing zero, show that there exists a branch of the logarithm.
Solution: As long as we can take the function $$ z \mapsto L(z) = \ln(|z|) + i \arg(z), $$ where $\ln$ is the $\log_e$ that one knows from calculus and one takes the following branch of $\operatorname{Arg}$ given by $\arg(z) = \{\theta \in [-\pi,\pi) : z = |z| e^{i \theta}\}.$ Then it sufficies to show that the function above is continuous. Indeed, $\ln$ and the absolute values are, so it is only left to show that $\arg$ is. By the definition of $\arg(z)$, one has $$ \arg(z) = \theta \iff x+iy = |x+iy| e^{i \theta} \iff x = r \cos(\theta) \text{ and } y = r \sin(\theta). $$ Using elementary geometry one then can calculate explicitly $$ \arg(x+iy) = \begin{cases} \arctan(y/x) &, x > 0, \\ \pi + \arctan(y/x) &, x < 0 \text{ and } y \geq 0, \\ -\pi + \arctan(y/x) &, x < 0 \text{ and } y < 0, \\ +\pi/2 &, x = 0 \text{ and } y > 0, \\ -\pi/2 &, x =0 \text{ and } y < 0. \\ \end{cases}$$ So in order to show that $\arg(\cdot)$ is continuous now it sufficies to take any sequence $(z_n)$ that converges to some $z \in D$ and show that $$ \lim_{n \to \infty} \operatorname{arg}(z_n) = \lim_{n \to \infty} \operatorname{arg}(x_n+iy_n) = \operatorname{arg}(x+iy) =\operatorname{arg}(z). $$ But using the rewriting and different cases from above this seems to hold true, since $\arctan$ from above seems continuous but now we have not used the simply connectedness yet, so I am missing something obvious? (Note: My definition only needs continuity).
Thanks in advance!