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Let $S^1\subseteq\mathbb{C}$ be the unit circle and let $U\subseteq S^1$ be open. How to show that there exist a continuous function $$\theta:U\to\mathbb{R}$$ such that $$e^{i\theta(z)}=z$$ for all $z\in U$ if and only if $U\neq S^1$?

I have been able to show the $\Rightarrow$ direction but not the other one.

$(\Rightarrow)$ If such a continuous $\theta:S^1\to\mathbb{R}$ exist, then it is injective and so $\theta(S^1)$ is a subset of $\mathbb{R}$ that is homeomorphic to $S^1$. Now, $S^1$ is compact and connected so $\theta(S^1)$ is a closed interval $[a,b]$. But removing a point from $[a,b]$ gives two connected components, while removing a point from $S^1$ still gives a connected space. Hence, $[a,b]\not\cong S^1$, so $\theta$ doesn't exist.

How would you show the $\Leftarrow$ direction?

Edit: This is an exercise in topology with no assumed prior knowledge of complex analysis. Note that we only need to show continuity of $\theta$ (not analyticity). The question is phrased in terms of complex numbers for simplicity of notation, but it is to be interpreted in terms of subspace of $\mathbb{R}^2$.

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    $\begingroup$ How did you prove the injectivity of $θ$? Furthermore, how did you conclude that $θ\colon S^1 → θ(S^1)$ is not only a bijective continuous map, but a homeomorphism? $\endgroup$
    – k.stm
    Jan 18 '15 at 14:59
  • $\begingroup$ By the way, it should read “for all $z ∈ U$”. $\endgroup$
    – k.stm
    Jan 18 '15 at 15:02
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    $\begingroup$ @k.stm It is injective because $\theta(z_1)=\theta(z_2)$ implies $z_1=e^{i\theta(z_1)}=e^{i\theta(z_2)}=z_2$. It is a homeomorphism because $S^1$ is compact and $\mathbb{R}$ is Hausdorff. $\endgroup$
    – Steve
    Jan 18 '15 at 15:04
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    $\begingroup$ Late to the party, but you can do this easier by defining $g(z)=f(z)-f(-z)$ and using the IVT to show that there is a point $z$ on the circle such that $f(z)=f(-z).$ $\endgroup$ Oct 11 '19 at 15:12
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You have actually proved the hard direction. The other direction is easy: if $U$ is not all of $S^1$, it misses at least one point $z\in S^1$. Without loss of generality, this point is $1=e^0$. Then define $\theta(e^{it})=t$, where $0<t<2\pi$. This function is continuous at every point of $S^1$ except $1$.

Proof. Suppose that $z_n=e^{i t_n}$ converges to $z=e^{it}\in S^1 -1$, but that $t_n$ doesn't converge to $t\in (0,2\pi)$. Since $t_n\in (0,2\pi)$, $t_n$ must have a convergent subsequence to some $t^*\in [0,2\pi]$ different from $t$. Note $0<|t-t^*|<2\pi$. Since $e^{iz}$ is continuous, we must have that $e^{i(t-t^*)}=e^0=1$. But $z=2\pi$ is the least nonzero real for which $e^{iz}=0$, so we've arrived at a contradiction.

I will provide a proof of a related fact. Try to see if you can modify it to prove what you want: suppose $V$ is an open subset of $\Bbb C$ and $f$ is a branch of the logarithm in $V$. (By definition, this is a continuous function $f:U\to \Bbb C$ for which $e^{f(z)}=z$. One can show that this guarantees $f$ is (complex) differentiable.) Then $S^1\not\subseteq V$.

Proof. Suppose $S^1\subseteq V$. Since $e^{f(z)}=z$, $f'(z)=1/z$. But then $$0= \frac{1}{2\pi i}\int_{S^1} f'(z) dz = \frac{1}{2\pi i}\int_{S^1} \frac{dz}z=1$$

Note that if you had a continuous map $u:S^1\to\Bbb R$ for which $e^{i u(z)}=z$, we would be able to define $f:\Bbb C^\times \to\Bbb C$ by $f(z)=\log |z|+ u(z|z|^{-1})$. This is continuous, and would furnish a branch of the logarithm in all of $\Bbb C^\times$, which is an open set containing $S^1$.

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  • $\begingroup$ Please look at what is written in the Edit section that I made a few hours ago. @MikeMiller is right; I have absolutely no background in complex analysis, so this answer is not useful to me. $\endgroup$
    – Steve
    Jan 18 '15 at 23:02
  • $\begingroup$ @Steve Well, it will be useful to other people consulting this post. I do hope it will be useful to you once you read some complex analysis. The general idea is that branches of the logarithm cannot be defined on open sets that have loops circling the origin. $\endgroup$
    – Pedro Tamaroff
    Jan 18 '15 at 23:05
  • $\begingroup$ Why is your function $\theta:S^1-\{1\}\to\mathbb{R}$ continuous? $\endgroup$
    – Steve
    Jan 19 '15 at 2:25
  • $\begingroup$ @Steve Let me edit that in. $\endgroup$
    – Pedro Tamaroff
    Jan 19 '15 at 2:27
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Let $\theta (z)=\Im (\log z)$ where $\log z$ is the branch for which $0<\arg z<2\pi$. This should work unless I'm missing something.

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  • $\begingroup$ You are missing $U$ and I should have clarified that this is an exercise in topology with no assumed prior knowledge of complex analysis. $\endgroup$
    – Steve
    Jan 18 '15 at 16:31
  • $\begingroup$ You should add this to the question, it seems to me that there is no way to do it without using some basic complex analysis facts. $\endgroup$
    – Zero
    Jan 18 '15 at 17:13
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For the other direction, take $z_0 \in \mathbb S^1\setminus U \not= \emptyset$.Then, for some $t_0 \in \mathbb R$ $$z_0 = e^{it_0}$$ Define $$\theta(z) :=\arg(z e^{-it_0}) + t_0$$ where $\arg$ takes values in $[0,2\pi)$. $\arg$ is discontinous in $1$, but if $z\in U$ then $z e^{-it_0} \not =1$ so $\theta$ is continous in $U$ and you can easily verify the condition.

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