Rigorous way to show image of a set under rotation 
Take $R:\mathbb{C}\rightarrow\mathbb{C}$ where
  $$R(z)=ze^{i\frac{\pi}{4}}.$$
  Find $R(A)$ where
  $$A=\{ re^{i\theta}:r\in [0,2], 0\leq\theta\leq\pi\}.$$

I've ran in to a sort of problem (it seems) with the way that this set is defined. I haven't done a lot of complex analysis. It looks to me like say
$$e^{i\pi} \in A \text{ but } e^{i3\pi} \notin A$$
as $3\pi \notin [0,\pi]$ but $e^{i3\pi}=e^{i\pi}$ so $e^{i\pi} \in A$ and $e^{i\pi} \notin A$.
To work through this problem (I made up the problem as an exercise) mathematically I made an equivalence relation with saying $a \sim b$, $a,b \in \mathbb{R}$ if $a-b=2\pi n$ for some $n \in \mathbb{Z}$. Then I defined the set as
$$A'=\{ re^{i\theta}:r\in [0,2], \theta \sim \alpha, 0\leq \alpha \leq \pi \}.$$
Is there a nice way to do this problem with some kind of mathematical rigor? Nicer than this at least! I know the answer would just be
$$R(A')=\{ re^{i\theta}:r\in [0,2], \theta \sim \alpha, \frac{\pi}{4}\leq \alpha \leq \frac{5\pi}{4} \}$$
unless there is another thing I am overlooking.
 A: Edit: When we say that $A=\{re^{i\theta}:r\in[0,2],0\le\theta\le\pi\},$ that is a bit misleading, and indeed, it looks like the set wouldn't even be well-defined! Ultimately, we are treating $A$, itself, as a function image. In particular, define $f:\Bbb R^2\to\Bbb C$ by $f(r,\theta)=re^{i\theta}.$ Then letting $B=[0,2]\times[0,\pi],$ all we're saying is that $A=f(B).$ It's certainly true that $f(1,3\pi)\in A,$ even though $\langle 1,3\pi\rangle\notin B,$ but that's nothing to worry about.
If we were being more careful, we would say $$A=\left\{z\in\Bbb C:\exists r\in[0,2],\theta\in[0,\pi]\text{ with }z=re^{i\theta}\right\}.$$
Take any $z\in A$. This means that $z=re^{i\theta}$ where $0\le r\le2$ and $0\le \theta\le\pi.$ Then $$R(z)=ze^{i\frac{\pi}4}=re^{i\theta}e^{\frac{i\pi}4}=re^{i\left(\theta+\frac{\pi}4\right)},$$ and so $|R(z)|=r\in [0,2]$ and the principal argument of $R(z)$ is in the interval $[\frac\pi4,\frac{5\pi}4].$ The reverse inclusion holds as well, so that $R(A)$ is precisely the set $R(A')$ (as you defined it). In our "more careful" notation (doing away with the equivalence relation), we have $$R(A)=\left\{z\in\Bbb C:\exists r\in[0,2],\theta\in\left[\frac\pi4,\frac{5\pi}4\right]\text{ with }z=re^{i\theta}\right\}.$$
