# Showing that $e^{ix}$ covers the whole unit circle

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. –  user20266 May 29 '12 at 17:41
Following @Thomas, this means you only need to choose the sign of $x$ to finish. –  copper.hat May 29 '12 at 17:50
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) –  M Turgeon May 29 '12 at 17:55
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 $x$. –  M Turgeon May 29 '12 at 17:56
@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? –  Nga May 29 '12 at 18:49
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.