Maximal simply connected open subset of $\mathbb{C}^*$

Question

Is it true that if A is a set maximal with respect to the property

" $ A\subseteq \mathbb{C}^*$ open subset of $\mathbb{C}^*$ and A simply connected"

then there exists an injective curve $\gamma:[0,+\infty)\rightarrow\mathbb{C}$ such that $\gamma(0)=0, \gamma(t)\rightarrow\infty$ for $t\rightarrow +\infty$, $\gamma$ is differentiable almost everywhere and such that $A=\mathbb{C}$\ Im($\gamma$)?

Intuition suggests me so, but it could well be that, for example, there is a $C^0$ curve $\gamma$ fulfilling all conditions above, except for the differentiability property.

I am interested in knowing what form can sets with such property have because they are the domain of definition of a continuous local section of the covering map $\mathbb{C} \rightarrow\mathbb{C}^*$,$z\mapsto e^z$, i.e a continuous (and holomorphic) branch of the logarithm.

Answer 1

I think I see now that my conjecture was false. Indeed, if f is the Weierstrass function as in https://en.wikipedia.org/wiki/Weierstrass_function, let $\gamma(t)=t+i(f(t)-f(0))$. Then $\gamma$ is continuous, nowhere differentiable, tends to $\infty$ as $t\rightarrow\infty$, and is injective (since $\gamma(t)=\gamma(t')$ $\Rightarrow$ t=t', looking at real parts). Moreover, if A=$\mathbb{C}$\Im($\gamma$), then A is open, simply connected and maximal with respect to these properties. It remains to check that Im($\gamma$) cannot be the image of a piecewise $C^1$ function, but this is true, otherwise Im($\gamma$) would be locally (excluding a discrete number of points) a $C^1$ one-dimensional manifold, but this is not true for this $\gamma$ (it should be proved, but it is a bit technical, I believe).