$e^{i\theta_n}\to e^{i\theta}\implies \theta_n\to\theta$ 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.

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 $ . 
