What would a graph of a function $f: \mathbb{R}^2 \to \mathbb{C}$ look like if $$f(x,y) = x^y$$
This is a question that's been on my mind since I was first introduced to exponential functions. I know that negative bases have odd roots, such as $(-1)^\frac{1}{3}=-1$, but aren't there an infinite amount of rationals with odd and even denominators between any two real values? Then there would be an infinite amount of points with real valued solutions and complex solutions right next to each other when $x<0$ and $y$ is on some interval $a \leq y\leq b$. So there would be points on the graph of $x^y$ which are discontinuous, but still very close together. So what would this graph look like? How would you intuitively understand the graph?
In order for $f(x, y) = \exp(y \log x)$ to be well-defined you have to fix a branch of $\log$ on the real axis. A standard choice is $$ \log x = \begin{cases} \ln x & x > 0, \\ \ln(-x) + i\pi & x < 0. \end{cases} $$ With this definition, $$ f(x, y) = \exp(y \log x) = \begin{cases} x^{y} & x > 0, \\ (-x)^{y} \exp(i\pi y) & x < 0. \end{cases} $$
Let $u + iv$ denote the standard complex coordinate on the image plane. When $x > 0$, you get the graph of the real-valued function $f(x, y) = x^{y}$, i.e., $$ u = x^{y},\quad v = 0. $$ If $x < 0$, you instead get the graph of a complex-valued function: $$ u = (-x)^{y} \cos(\pi y),\quad v = (-x)^{y} \sin(\pi y). $$ Geometrically, take the graph of the real-valued function $g(x, y) = (-x)^{y}$ and rotate each "strip" over $y = \text{const}$ by an angle $\pi y$ in the $(u, v)$-plane.
Parametrically, the graph is $$ (x, y) \mapsto \begin{cases} (x, y, x^{y}, 0) & x > 0, \\ \bigl(x, y, (-x)^{y} \cos(\pi y), (-x)^{y} \sin(\pi y)\bigr) & x < 0. \end{cases} $$ You can attempt to plot this by projecting away a coordinate, or by otherwise combining two coordinates to get a third, or using a color density to represent one coordinate. (I tried a few obvious techniques along these lines, but didn't find a surface that leaped out as especially useful geometrically.)