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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?

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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
up vote 6 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

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


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That's neat, thanks! – Dirk Dec 13 '11 at 16:43
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

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.

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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
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

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