Existence of continuous angle function $\theta:S^1\to\mathbb{R}$ 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 $\mathbb{R}^2$.
 A: 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 $S^1$. 
A: 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.
A: 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.
