Let $\alpha>1$ be a real number, and consider the map $F$ defined on the closed unit two-dimensional disk $D \subseteq \mathbb{R}^2$ after removing a ray, by $re^{i\theta} \to re^{i\alpha\theta}$. Does $F \in W^{1,2}(D,\mathbb{R}^2)$?
When $\alpha=n$ is a positive integer, I proved that the answer is positive (see below, I hope I don't have a mistake).
I am not sure what to do when $\alpha \notin \mathbb{N}$.
The main difference is that when $\alpha=n$ is a positive integer, we have that $F(z)=F(re^{i\theta})= re^{in\theta}=\frac{z^n}{|z|^{n-1}}$ is defined and smooth on the entire disk without the origin $D \setminus \{0\}$. However, when $\alpha \notin \mathbb{N}$, we have to use the complex logarithm in order to define $F$, so (I think) we only get a well-defined map which smooth on the disk with a ray removed from it.
Here is the computation that shows that $F \in W^{1,2}$ when $\alpha=n$ is an integer:
$F(z)=\frac{z^n}{|z|^{n-1}}=\big( \frac{z}{|z|^{\frac{n-1}{n}}}\big)^n$ is the $n$-th power of a continuous bounded function defined on the entire disk, and smooth on $D \setminus \{0\}$. Thus, it suffices to show that $z \to \frac{z}{|z|^{\frac{n-1}{n}}}$ has a derivative which is in $L^2$.
Writing $\beta:=\frac{n-1}{n}$, we need to differentiate $g(z)=\frac{z}{|z|^\beta}=(\frac{x}{(x^2+y^2)^{\frac{\beta}{2}}},\frac{y}{(x^2+y^2)^{\frac{\beta}{2}}})$.
By symmetry, it suffices to check $G(x,y)=\frac{x}{(x^2+y^2)^{\frac{\beta}{2}}}$.
$G_x=(x^2+y^2)^{-\frac{\beta}{2}}[1-\beta \frac{x^2}{x^2+y^2}]$, so (since $0<\beta<1$), we get $|G_x| \le (x^2+y^2)^{-\frac{\beta}{2}}=r^{-\beta}$.
Thus $|G_x|^2 \le r^{-2\beta}$, so $|G_x|^2 \in L^1$ iff $ \int_0^1 r^{-2\beta} rdr < \infty$. Since $-1<1-2\beta=\frac{2-n}{n} \le 0$, this holds.
Similarly, $|G_y|=|(x^2+y^2)^{-\frac{\beta}{2}}\frac{2xy}{x^2+y^2}| \le (x^2+y^2)^{-\frac{\beta}{2}}$, so we are done.
In polar coordinates on the domain, your map is $$ (r,\theta)\mapsto (r\cos(\alpha\theta),r\sin(\alpha \theta)).$$ Note that since you removed a ray from your domain, polar coordinates form a global coordinate system.
We therefore have $$ |dF(p)|^2 = \partial_i F^\alpha(p) \partial_j F^\beta(p) g^{ij}(p) \delta_{\alpha\beta} = 1 +\alpha^2,$$ where $\delta_{\alpha\beta}$ is the Euclidean metric in Cartesian coordinates, and $g^{ij}$ is the inverse of the Euclidean metric in polar coordinates (that is, $g^{11} = 1$, $g^{12}=0$, $g^{22} = r^{-2}$).
It is immediate that this has a finite integral, hence your map is in $W^{1,2}$.
Note, however, that without removing a ray from the domain, the map has a discontinuity on the ray for non-integer $\alpha$, hence it cannot be in $W^{1,2}(D;R^2)$ (this space contains functions that are not continuous, but the set of discontinuity cannot be so large).