Let $D\subset\Bbb C$ be the complement of the closed spiral $\{e^{\theta+i\theta}:\, \theta\in\Bbb R\}\cup\{ 0\}$. Let $\log$ be a branch of the logarithm in $D$ such that $\log e=1$. Find $\log e^{15}.$
I am always confused in these type of logical concept because what I am feeling that no branch can exist as $D$ is region around zero. So the argument function will be discontinuous. But there is a catch that the spiral is dense at zero, so possibly we are not getting a region around zero. Basically if I just see on a naked eye then the value should be $15$. How to write the answer with proper logic.
I saw one related question here but got no answers and even the try was hard to understand.
A few notes about a branch of the logarithm first:
A branch of the logarithm is a continuous function $L$ defined on a simply connected domain $D$ such that $e^{L(z)} = z$ for all $z$.
Since $e^z \neq 0$ for all $ z$ we cannot have $0 \in D$.
We have that $e^z = e^w$ iff $z=w+2\pi n i$ for some $n$.
If $L_1,L_2$ are two branches of the logarithm defined on $D$, then we must have $L_1(z) = L_2(z) + 2 \pi n_z i$, and since $L_k$ are continuous, we must have $L_1(z) = L_2(z) + 2 \pi n i$ for some fixed $n$ and for all $z \in D$.
Please help from here!