Determine the set of values of $\exp(1/z)$ for $0<|z|<r$

Question

I can't do this exercise of Conway's Book: For $r>0$ let $A=\{w:w=\exp(1/z), 0<|z|<r\}$, determine the set $A$. Any hints?

Answer 1

Let's prove that $A = \mathbb{C} - \{0\}$. Let's pick $r > 0$. First, we'll prove that $A \subset \mathbb{C} - \{0\}$, then we'll show that $\mathbb{C} - \{0\} \subset A$.

1. Let's show that $A \subset \mathbb{C} - \{0\}$. This amounts to showing that $0$ does not belong to A. Given any $w, z$ such that $w = exp(1/z)$ and $0 < |z| < r$, we can write: $1/z = a + ib$ (with $a, b \in \mathbb{R}$). Then $|w| = e^a|e^ {ib}| > 0$. Thus, $0$ is not in A.
2. Now, let's show that any non-zero complex number $x$ can be written as $x = \exp(1/z)$ with $0 < |z| < r$. This will demonstrate that $A \subset \mathbb{C} - \{0\}$. For that, we pick $x \neq 0$. We write $x$ in exponential form ($\rho, \theta \in \mathbb{R}$): \begin{eqnarray} x & = & \rho e^{i\theta}\\ & = & \exp(ln(\rho) + i\theta) \end{eqnarray} Let's pick $k \in \mathbb{N}$ such that $|ln(\rho) + i\theta + 2k\pi| > 1/r$. Then $z = ln(\rho) + i\theta + 2k\pi$ verifies $0 < |z| < r$ and we have $x = exp(1/z)$. Thus $x$ belongs to $A$.

We have shown that A both contains and is a subset of $\mathbb{C} - \{0\}$. Thus $A = \mathbb{C} - \{0\}$

Answer 2

$\exp$ maps each horizontal strip of height $2\pi$ onto $\mathbb{C}\setminus \{0\}$, and for all $r\gt0$, $\{w: w=\frac{1}{z},0\lt|z|\lt r\}$ contains infinitely many such strips.

Answer 3

The Big Picard Theorem (http://en.wikipedia.org/wiki/Picard_theorem) is a great generalization of Casorati-Weierstrass which says that in the neighborhood of an essential singularity a holomorphic function assumes all complex values with possibly one exception. In this case, that is clearly $0$. So the set is $\mathbb{C}-\{0\}$.