Image of a small open disk under $f(z)= \exp (1/z)$ In Conway's Functions of One Complex Variable I, the function $f(z)= \exp (1/z)$ has an essential singularity at $z=0$. But what would the image of a small punctured open disk $\{z:0<|z|< \delta\}$ be under this function? I'm having a hard time visualizing it.
 A: Take $z = \epsilon e^{i \theta}$ to be an arbitrary point in the open disk (with $0 < \epsilon < \delta$).  We note that
$$
\exp(1/z) = \exp(\epsilon^{-1} \exp(-i\theta)) = 
\exp(\epsilon^{-1} [\cos \theta - i\sin \theta]) =\\
\exp[\epsilon^{-1}\cos \theta]\exp[-i\epsilon^{-1}\sin \theta]
$$
Now, suppose we have an arbitrary non-zero complex number $z' = re^{i \theta'}$.  Then $z'$ is in the image of the disk if there is a solution to the system
$$
\cos \theta = \epsilon \ln(r)\\
\sin \theta = -\epsilon[\theta' + 2 \pi k]\\
\epsilon < \delta, \quad k \in \mathbb{Z}
$$
That is, $z'$ will be in the image of the disk as long as we can select an $\epsilon < \delta$ and $k \in \mathbb{Z}$ such that
$$
\epsilon^2\left([\ln(r)]^2 + [\theta' + 2\pi k]^2\right) = 1
$$
From here, it should be clear that every non-zero $z'$ will be in the image.  In particular, we may prove it as follows:
Fix $k > 1/\delta^2$.  It is clear that $\left([\ln(r)]^2 + [\theta' + 2\pi k]^2\right) > 1/\delta^2$.  Define 
$$
f(\epsilon) = \epsilon^2\left([\ln(r)]^2 + [\theta' + 2\pi k]^2\right)
$$
We note that $f(0) = 0 < 1$, and $f(\delta) > 1$.  By the intermediate value theorem, $f(\epsilon) = 1$ has a solution with $\epsilon \in (0,\delta)$.
