The function $e^{iz}$ maps $[0,2\pi)$ bijectively to $\{z\in\Bbb C:|z|=1\}$ Does anyone know a good proof or reference for this fact? I'm trying to write an elementary formal proof of this from just the series for $e^z=\sum_{n=0}^\infty z^n/n!$ (I don't have calculus available yet). So far, I have the basic trigonometric facts about $\pi$ and $e^{iz}$ and $\sin x,\cos x$; this means things like $e^{x+y}=e^xe^y$, $\sin^2x+\cos^2x=1$, $e^{i\pi}=-1$, $\sin(x+2\pi n)=\sin x$, $\sin(x+\pi/2)=\cos x$, addition formulas for $\sin$ and $\cos$, etc.
Let $f(x)=e^{ix}$ for $x\in[0,2\pi)$. It is easy to show that $$|f(x)|=|\cos x+i\sin x|=\cos^2 x+\sin^2x=1,$$ so $f$ is a function into $\{z\in\Bbb C:|z|=1\}$. If $e^{ix}=e^{iy}$, then $e^{i(x-y)}=1$, so it is sufficient to prove that $e^z=1$ iff $\frac z{2\pi i}\in\Bbb Z$. I have a proof that $\sin z=0$ iff $\frac z\pi\in\Bbb Z$ already, and it follows from this and $\cos 2z=1-\sin^2 z$ that $\cos z=1\iff\frac z{2\pi}\in\Bbb Z$, but I don't quite see yet how to make the leap to $e^{iz}$ from this information.
For surjectivity, I have the intermediate value theorem available, so there is a $z\in(0,\pi)$ such that $\cos z=x$ for any $x\in(-1,1)$. Is there a trick to avoid the case analysis here for the various combinations of $z$ in each quadrant and on the axes?
 A: I'll use some properties of arctangent.
First, from $\cos^2 x+\sin^2 x=1$, you have $|e^{it}|=1$.
You may use periodicity: since $\cos$ and $\sin$ are $2\pi$-periodic, the function can't be bijective on a larger interval.
Now, if $t, t' \in [0,2\pi[$, and $e^{it}=e^{it'}$, then $e^{i(t-t')}=1$, so, with $\eta=t-t'$, $e^{i(z+\eta)}=e^{iz}e^{i\eta}=e^{iz}$ so $\exp$ is $\eta-periodic$. But since $2\pi$ is the smallest period (of $\sin$ and $\cos$), you must have $\eta=2k\pi$ for some integer $k$, contradiction.
Thus, for different values of $t\in[0,2\pi[$, values of $e^{it}$ must be different, that is the function is injective.
But we have yet to prove surjectivity.
If $|z|=1$, for $z=x+iy$, then $x^2+y^2=1$, and suppose also $x>0$ and $y>0$, and let $t=\arctan \frac{y}{x}$.
You have $\frac{1}{\cos^2 a} = 1+\tan^2 a$ so $\cos a=\frac{1}{\sqrt{1+\tan^2 a}}$ and specifically, $\cos t=\frac{1}{\sqrt{1+\frac{y^2}{x^2}}}=\frac{x}{x^2+y^2}=x$. You can prove the same way that $\sin t=y$.
A similar method will hold for other values of $x$ and $y$ (with negative of zero values).
Hence your function is also surjective, so it's bijective.

Regarding the cases for each quadrant, there is a nice formula for principal argument of a number $z=x+iy \notin \Bbb R_-$, it's
$$Arg (x+iy)=2\arctan \frac{y}{x+\sqrt{x^2+y^2}}$$
So maybe, using cleverly trigonometric formulas you can have mainly only one case.

There is an easy way to prove $2\pi$ is the smallest period of $\sin$, $\cos$.
From $e^z=\sum_{n=0}^\infty \frac{z^n}{n!}$ you prove $e^ze^{z'}=e^{z+z'}$ using a Cauchy product. Then you define $\sin$ and $\cos$ and you have most trigonometric formulas.
Then, define $\frac{\pi}2$ as the smallest positive root of $\cos$. There is at least one root since $\cos 0=1>0$, and $\cos 2<0$ (use majoration from the series). And this root is the smallest since the derivative of $\cos$ is $-\sin$, which is strictly negative on $]0,2]$.
Then you can prove $2\pi$ is a period of $\sin$, $\cos$ (thus $2i\pi$ a period of $\exp$). Using an argument on the roots, there is no smaller (real) period or $\sin$ and $\cos$.
And there is of course no other complex period (of $\sin$, $\cos$) either, because if $e^{z+i\,\eta}=e^z$, then $e^{i\eta}=1$ so if $\eta=x+iy$, you have $e^{-y+i\,x}=e^{-y}e^{i\,x}=1$ and its modulus is $e^{-y}=1$ so $y=0$.
