# Determine the set $\{w:w=\exp(1/z), 0<|z|<r\}$

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?

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Start by looking at the set {1/z: 0<|z|<r}, then think Euler's formula... –  hardmath Jan 9 '11 at 18:52
Another way to look at this is to notice that 0 is an essential singularity of the function, then use the Casorati Weierstrass theorem. –  user3180 Jan 9 '11 at 19:25
It can't be all of the complex plane since 0 is never in the range of the exponential function. However, it is a dense subset of the complex plane by CW theorem. –  user3180 Jan 9 '11 at 19:41
Is it all the complex plane minus the origin? Because of the periodicity of the exponential. (Obs.: Sorry, I meant the complex plane minus the origin, I've excluded the previous message). –  Jacques Jan 9 '11 at 19:45
I think you're right. $A$ is also equivilent to {$w:w=exp(z)$, $|z|>$$1/r}. The exponential function maps C onto C-\{0\}. You can then use periodicity to argue that if |z|<$$ 1/r$ then there exists $t$ such that $|t|>$$1/r$ and exp(z)=exp(t). Hence, the range or your function is the same as the range of the exponential function on all of $C$. Namely,$C-\{0\}$. The interesting thing is that this is true no mater how small $r$ is. That's the point of the Casorati Weierstrass theorem. –  user3180 Jan 9 '11 at 20:15

$\exp$ 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.

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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 $\mathbb{C}-\{0\}$.

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The "Big Picard Theorem" is a very deep theorem about essential singularities of arbitrary holomorphic functions, and you will not see a proof of it in an undergraduate course. But here we have a very special function whose behavior we can describe in an elementary way. So there is no need to appeal to such heavy machinary. Just look at, e.g., the answer of Jonas Meyer. –  Christian Blatter Jan 12 '11 at 19:14

Let's prove that $A = \mathbb{C} - \{0\}$. 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 z such that $0 < |z| < |r|$ then we can write: $1/z = a + ib$ (with $a, b \in \mathbb{R}$). Then $|z| = exp(a) > 0$. 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 $A = \mathbb{C} - \{0\}$

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