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In my book analysis they argue that $e^{ix}$ covers the whole unit circle as follows.

Suppose that $w = u +iv$, such that $|w| = 1$. Then if $v \geq 0$ we pick $x \in [0,\pi]$ with $\cos x = u$. else if $v < 0$ we pick $x \in [\pi,2\pi]$, with $\cos x = u$.

I really do not see why this works. I know we have to show that there exists $x$ and $y$ such that $\cos x = u$ and $\sin y = v$. Furthermore I know that due to the intermediate value theorem, since cosine is continuous on $[0,\pi]$ that if $u$ is in between $\cos 0=1$ and $\cos \pi = -1$, there exists $x \in ]0,\pi[$ such that $\cos x = u$.

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    Note, that if you find $x$ as indicated, $\sin^2 (x) + \cos^2(x) = 1$ -- this is always true.2012-05-29
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    Following @Thomas, this means you only need to choose the sign of $x$ to finish.2012-05-29
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    You do not want $x$ and $y$; you only want some $x$ such that $e^{ix}=w$. Knowing that $|w|=1$ ensures that the $x$ you choose will also satisfy $\sin x=v$ (as noted by @Thomas)2012-05-29
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    Actually, to put another way, you already know that $w=\cos x+i\sin x$ for some $x$ (this is polar coordinates). Hence, $w=e^{ix}$ for the same $x$.2012-05-29
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    @Thomas, Ah I think I know now, for $x \in [0,\pi]$, we have $\sin x$ is positive. Since $u^2 + v^2 = 1$, and since we have an $x$ such that $\cos x = u$, and since we have $\sin^2 x + \cos^2 x = 1$, we have $1-\cos^2 x = \sin^2 x = 1-u^2 = v^2$, hence $\sin^2 x = v^2$ and thus since we choose $v$ and $\sin x$ with the same sign, we have $\sin x = v$. Is this correct?2012-05-29
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    @Nga: that's the general idea, yes. You have to ensure you get the whole circle, though, so a bit more of work is required.2012-05-29
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    @Thomas, could I ask what you think is missing then? Since we chose $w$ as an arbitrary point on the circle I don't really see an issue.2012-05-29
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    @Nga It's just the coverage of all of the circle. You mention $x\in [0,\pi]$, this covers at most half of it.2012-05-29
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    @Thomas, oh ok thanks.2012-05-29

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