If we consider $f(z) = \exp\left(\frac{1}{z^2}\right)$, how do we compute $f(D^*(0,r))$, where $D^*(0,r) = \{z\in\Bbb C : 0<|z|<r\}$?
We know that $f$ has an essential singularity in 0, so by Cassorati-Weierstrass' theorem we know that the image of any perfored neighbourhood of 0 is dense in $\Bbb C$. Does this mean that $f(D^*(0,r)) = \Bbb C$?
If this is not important, how should I approach this problem instead?
I'm putting the discussion in the comments into an answer in order to have it more clean and mark the question as answered.
Consider first $D_1 = D^*(0,r) = \{z\in\Bbb C : 0<|z|<r\}$ and $f_1(z) = z^2$, and we evaluate $f(D_1) = D^*(0,r^2) = \{z\in\Bbb C : 0<|z|<r^2\}=:D_2$. Now we take $f_2(z) = \frac{1}{f_1(z)} = \frac{1}{z^2}$, and get $f_2(D_2) = \Bbb C\setminus D(0,r^2) = \{z\in\Bbb C : |z|>r^2\}=:D_3$.
Since $\exp$ is $2\pi i$-periodic, it maps each horizontal strip of height $2\pi$ onto its total image. We only need to check that $D_3$ contains a horizontal strip of height $2\pi$, but this is simple (take for example $S = \{z\in\Bbb C : 2r^2< \Im(z)<2r^2+2\pi\}$, which is the horizontal strip of height $2\pi$ and starts at $2r^2$, in order to contain it in $D_3$).
Therefore, $f(D_1) = \Bbb C$.
This could further be revised with the Cassorati-Weierstrass theorem, since $\exp\left(\frac{1}{z^2}\right)$ has an essential singularity at $z=0$. Therefore, the image of $D^*(0,r)$ (a perfored neighbourhood of $0$) is dense.
I want to express my thanks to Andrew D. Hwang for assisting me in this problem. Sorry if I felt annoying!