How do I draw the image of the following complex region under the power map? 
Let me denote $R=\left\{0<Arg(z)<\frac{\pi}{6}\right\}$. I want to understand how $f(R)$ looks like if $f(z)=z^i$.

My idea was that I first compute the image for an arbitrary $z\in R$ and then try to draw it.
So let $z\in R$ then $z=re^{i\theta}$ where $r\in (0,\infty)$ and $\theta \in \left(0,\frac{\pi}{6}\right)$. Then $$f(z)=f(re^{i\theta})=e^{i(\log(r)+i\theta)}=\frac{1}{e^\theta}e^{i\log(r)}$$ Where $\theta\in \left(0, \frac{\pi}{6}\right)$ and $\log(r)\in (0,\infty)$
So now our new radius is $\frac{1}{e^\theta}$ and the new argument is $\log(r)$.
But this somehow seems a bit strange to me, is this really correct? And if yes how do I draw it now from here?
My idea is that the new radius let's denotes it $r'$ satisfies $\frac{1}{e^{\frac{\pi}{6}}}<r'<\frac{1}{e^0}=1$ but then the argument lies in $(0,\infty)$?
 A: Per request for mapping help:
Each ray (i.e. half line, emanating from the origin) of direction $\theta$ goes to a concentric circle of radius $e^{-\theta}.$
Along this ray, as $r$ goes from $0 < r$ to $r < \infty$, the argument $\log(r)$ will go from $-\infty$ to  $+\infty$.  Strangely, this is
of only arguable relevance, because it is obvious that you can find a subset of $r$ : $R_1 \leq r < R_2$ such that $\log(R_2) - \log(R_1) = 2\pi$.
This means that as $r$ takes on the values in $[R_1, R_2)$ that a complete revolution is traced out, along the circle of radius $e^{-i\theta}.$
So, the map should be represented by a continuous series of concentric circles, which is represented by the closed donut.  The inner radius of the donut is $e^{(-\pi/6)}$ and the outer radius is $e^{-0}$.

An alternate perspective is that if you hold $r$ fixed, with $\alpha = \log(r)$, and let $\theta$ take on the values in $[0,\pi/6]$, that a ray will be traced along the path $\alpha$, from radius $e^{(-\theta/6)}$ to $e^{(-0)}.$
