Let $f: \mathbb{C} \to \mathbb{C}$ be a function given by $f(z) = e^z$. How would you parametrise the boundary of the image $f(B(a,r))$ ($\partial f(B(a,r))$) with $B(a,r) \subseteq \mathbb{C}$ being a sphere around $a \in \mathbb{C}$ with radius $r > 0$?
My hunch is that the boundary $$ f(\partial B(a,r)) = \partial f(B(a,r)) $$ and I thought that would make it easier, but I couldn't prove it (because maybe it isn't even true). In that case it would just be $$ \gamma(\varphi) = e^{r e^{i \varphi}} $$ with $\varphi \in [0, 2 \pi)$. Furthermore, I tried messing with the fact that $$ |e^z| = e^{|z| \cos(\arg(z))} $$ and $$ \arg(e^z) = |z| \sin(\arg(z)). $$ Are there any tips/strategies? Thanks in advance!
In general, $f(\partial B(a,r)) \not= \partial f(B(a,r))$.
Indeed, for $a=0$ and $r=4$, the boundary of $B(a,r)$ is mapped to a curve that looks like a limaçon and has self-intersections. The self-intersections occur for $t=\theta$ and $t=2\pi-\theta$, where $\theta=\arcsin(\pi/4)$. These are found by solving $Im(f(z))=0$.