Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

An argument function $\phi$ on $\mathbb{C}\setminus\{0\} = \mathbb{R}^2\setminus\{0\}$ is a function such that for every $z\neq 0$ it holds that $$z = |z|\exp(i\phi(z)).$$

Is there an elementary and easy proof that there is no continuous argument function on $\mathbb{C}\setminus\{0\}$? I would like to see a proof which uses as less complex analysis as possible. Probably only topological arguments and no complex numbers whatoever?

share|improve this question
2  
No complex numbers? Could you please elaborate on what that would mean in the context of this problem? –  Jonas Meyer Dec 13 '11 at 15:19
add comment

2 Answers

up vote 5 down vote accepted

Since the question itself uses complex numbers, I don't think it's possible to give a proof that doesn't use them.

Anyway, the following is a simple argument that doesn't use any complex analysis.

Suppose such a function $\phi$ exists, and consider the function $\psi(t) = \frac{1}{2 \pi}(\phi(e^{i t}) - t)$, for real $t$. $\psi$ is continuous and $\mathbb{Z}$-valued, so it must be constant, say $\psi \equiv k$.

So $\phi(e^{it}) = 2 \pi k + t$, but then

$$ 2 \pi k = \phi(1) = \phi\left(e^{2 \pi i}\right) = 2 \pi (k + 1), $$

contradiction.

share|improve this answer
    
That's neat, thanks! –  Dirk Dec 13 '11 at 16:43
1  
The $\phi(2\pi)$ is too much, right? –  Dirk Dec 13 '11 at 17:25
    
Yep, sorry. Fixed. –  Paolo Capriotti Dec 13 '11 at 17:38
add comment

If there were a continuous argument function, its restriction to the unit circle would be a homeomorphism onto its image in $\mathbb R$. (It is injective, the circle is compact, and $\mathbb R$ is Hausdorff.) But the image of the circle would be a compact and connected subset of $\mathbb R$, thus a closed and bounded interval, which is not homeomorphic to the circle. This contradiction shows that such a function doesn't exist.

share|improve this answer
    
I had something like this in mind, thanks! However, this shifts the problem to the problems that the circle is not homeomorphic to a subset of the line which is true but also not very obvious to prove. –  Dirk Dec 13 '11 at 16:41
2  
Dirk, For one thing, note that it is not just any subset of the line, but an interval $[a,b]$ for some $a<b$. From $[a,b]\subset \mathbb R$ you can remove 2 points and have a connected set remaining, which is not true for the circle. Or, perhaps simpler, show that if $\varphi:S^1\to [a,b]$ is a continuous surjective map, $\varphi(z)=a$ and $\varphi(w)=b$, then $\varphi$ must map both of the arcs between $z$ and $w$ onto $(a,b)$, and therefore $\varphi$ is not injective. As in Paolo's argument, connectedness of $S^1$ is essential. (Or, the circle is not contractible.) –  Jonas Meyer Dec 13 '11 at 17:26
    
You are right. The argument is indeed not difficult. What I have not told is that I was looking for an argument to use in a lecture and I can not assume topological background. If I could, I would prefer your answer :-). –  Dirk Dec 13 '11 at 21:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.