$e^{i\theta_n}\to e^{i\theta}\implies \theta_n\to\theta$ How to show $e^{i\theta_n}\to e^{i\theta}\implies \theta_n\to\theta$ for $-\pi<\theta_n,\theta<\pi.$ I'm completely stuck in it. Please help.
 A: You can try to use the well-known (I think...) theorem:
Theorem: if $\,z_n=x_n+iy_n\in\Bbb C\,$  , then
$$\lim_{n\to\infty}z_n=z=x+iy\in\Bbb C\;\;\text{in}\;\;\Bbb C\Longleftrightarrow x_n\xrightarrow [n\to\infty]{} x\,\,\wedge\,\,y_n\xrightarrow[n\to\infty]{}y\,\,\;\;\text{in}\;\;\Bbb R$$
So
$$e^{i\theta_n}=\cos\theta_n+i\sin\theta_n\xrightarrow[n\to\infty]{}\cos\theta+i\sin\theta =e^{i\theta}\Longleftrightarrow$$
$$\cos\theta_n\xrightarrow[n\to\infty]{}\cos\theta\,\,\,\wedge\,\,\,\sin\theta_n\xrightarrow[n\to\infty]{}\sin\theta$$
And now use the continuity of the (real) trigonometric functions. For example:
$$\sin\theta=\lim_{n\to\infty}\sin\theta_n=\sin\left(\lim_{n\to\infty}\theta_n\right)\;\;\ldots$$
A: $$\Vert e^{i \theta_n} - e^{i \theta}\Vert_2^2 = (\cos(\theta_n) - \cos(\theta))^2 + (\sin(\theta_n) - \sin(\theta))^2 = 2-2\cos(\theta_n - \theta) \to 0$$
This implies $$\cos(\theta_n - \theta) \to 1 \,\,\,\,\,\,\,\,\, \text{(Why?)}$$
Can you now finish it off by noticing that $\theta_n ,\theta \in (-\pi,\pi)$?
A: Suppose $(\theta_n)$ does not converge to $\theta$, then there is an $\epsilon > 0$ and a subsequence $( \theta_{n_k}  )$ such that $| \theta_{n_k} - \theta | \geq \epsilon $ for all $k$.  $(\theta_{n_k})$ is bounded so it has a further subsequence $(\theta_{m_k})$ which converges to $\theta_0  \in [-\pi,\pi]$ (say) with $| \theta - \theta_0 | \geq \epsilon $, and hence $ \theta_0 \neq \theta $. Next $( \exp i\theta_{m_k} )$ being a subsequence of $( \exp i\theta_n ) $ must converge to $\exp i\theta $, hence $ \exp i\theta = \exp i\theta_0 $. So $ \theta_0 = 2n \pi + \theta $ for some integer $n$, however $ | \theta_0 - \theta | < 2\pi $ as $ \theta \in ( -\pi , \pi )$ and $ \theta_0 \in [-\pi,\pi]$, this implies $ \theta = \theta_0$ and contradicts $ \theta \neq \theta_0 $ . 
