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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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    @Thomas, oh ok thanks.2012-05-29

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Let's clean this up. We are given $(x,y)$ such that $x^2+y^2=1$. Since $\cos(0)=1$ and $\cos(\pi)=-1$, by the intermediate value theorem there exists $t\in [0,\pi]$ such that $\cos t=x$. Then $\sin t=\sqrt{1-x^2}=|y|$. So, if $y\ge 0$, this value of $t$ works. Otherwise, we use $-t$ (or $2\pi-t$), which has the same cosine but the sine of opposite sign.