1
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ Why has this question attracted so much trash? I personally think it's a really good question! (+1) $\endgroup$
    – user253055
    Nov 12, 2015 at 1:27

1 Answer 1

Reset to default
2
$\begingroup$

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.)

$\endgroup$
3
  • $\begingroup$ But $\forall k\in\mathbb{N}\quad\exp(y\log x) = (-x)^y \exp (i(2k+1)\pi y)$ so then for each $k$ a new graph could be made, correct? And then for any fixed $x<0$ I could say that $x^y$ is dense in $\mathbb{R}$ because it can be made arbitrarily close to any real valued number? $\endgroup$
    – cheesyfluff
    Nov 12, 2015 at 3:30
  • $\begingroup$ And would it be dense in $\mathbb{C}$? $\endgroup$
    – cheesyfluff
    Nov 12, 2015 at 3:46
  • 1
    $\begingroup$ Depending what you mean by $x^{y}$ (is it a function, or a relation?), "yes, you can choose another branch (or some/all branches) of $\log$." For fixed $x < 0$ and $k$, the image of each vertical line in the $(x, y)$ plane is a logarithmic spiral; the union of these spirals over all (integer) $k$ is indeed dense in $\mathbf{C}$, since if $y$ is irrational, the set of $\exp\bigl(i(2k + 1)\pi y\bigr)$ is dense in the unit circle. $\endgroup$
    – Andrew D. Hwang
    Nov 12, 2015 at 10:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.