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

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.

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Just curious... what do you mean by "$\to$" in this case? Is this a limit notation that I haven't come across yet in math? –  anorton Jan 10 '13 at 3:05
@anorton Typically, $f_n \to f$ implies $\lim_{n \to \infty} \Vert f_n - f \Vert = 0$. –  user17762 Jan 10 '13 at 3:06

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:

$$\sin\theta=\lim_{n\to\infty}\sin\theta_n=\sin\left(\lim_{n\to\infty}\theta_n\right)\;\;\ldots$$

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Inverse sine function is continuous everywhere $in$[-1,1} $\implies \theta_n \to \theta$ as $n\to \infty.$ Right? If it be the case, then where did we use the condition that $-\pi<\theta_n, \theta<\pi?$ –  Sriti Mallick Jan 10 '13 at 4:16
Yes to the first question, but about the second I think the image of $\,\arcsin\,$ is more usually chosen to be $\,[-\pi/2\,,\,\pi/2]\,$... –  DonAntonio Jan 10 '13 at 5:06
$[-1,1]\to[-{\pi},{\pi}]:x\mapsto2\sin^{-1}x$ is continuous on $[-1,1].$ So, $\sin\theta_n\to\sin\theta\implies2\sin^{-1}\sin\theta_n\to2\sin^{-1}\sin\theta_‌​n\implies\theta_n\to\theta.$ –  Sriti Mallick Jan 10 '13 at 11:14

$$\Vert e^{i \theta_n} - e^{i \theta}\Vert_2^2 = (\cos(\theta_n) - \cos(\theta))^2 + (\sin(\theta_n) - \sin(\theta))^2 = 2-2\cos(\theta_n - \theta) \to 0$$ This implies $$\cos(\theta_n - \theta) \to 1 \,\,\,\,\,\,\,\,\, \text{(Why?)}$$ Can you now finish it off by noticing that $\theta_n ,\theta \in (-\pi,\pi)$?

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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 $\theta \neq \theta_0$ .

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