complex numbers - How to visualize $f(x) = (-2)^x$

Background

I teach Algebra and second year Algebra to middle school students. We are currently studying Exponential, Power, and Logarithmic functions. We study exponential functions (of the form $f(x) = ab^x$ where $a \neq 0, b > 0$). I've taught the basics of exponential functions before and I've wondered about the function $f(x) = (-2)^x$. At first I was confused as to why the calculator would not graph any exponential function with a negative base. Then I realized that depending on the roots, say for $(-2)^{1.25}$ and $(-2)^{1/3}$ some very different calculations would need to be performed. I'm still lost as to how you would explain $(-2)^\pi$ but I suppose that's just being annoying. I want to know how to explain the output of $f(x) = (-2)^x$.

Comments

I know for example that there are a host of values for which the output would be imaginary. If we limit the domain of $x$ to rational numbers with odd denominators

Here are two examples

$(-2)^{2/5} = \left(\sqrt[5]{-2}\right)^2 \approx 1.3195$

$(-2)^{3/5} = \left(\sqrt[5]{-2}\right)^3 \approx -1.5157$)

we should be able to compute $f(x)$ for all such numbers. I would then think the graph would look something like

Curious how to visualize the plot including the imaginary outputs when $x \in \mathbb{R}$

Wolfram Computes the result using the principal roots

Question: How could you visualize and interpret $f(x) = (-2)^x$ while incorporating the real-root based output I presented above? Or, alternatively, what's wrong with this?

Wolfram|Alpha says that it has no properties as a real function which makes sense given the restricted domain necessary to compute real outputs. Is there any mistake in thinking that with the domain restrictions I specified above ($x$ all rational numbers such that $x = a/b$ where $b$ is odd) the output would resemble my plot above?

Is there a way to visualize -- or even better -- explain the behavior of this function for for all real numbers $x$? If we added in an imaginary axis, could we use it to show the results when we need to compute an imaginary output?

Answer

The following is a variation of the visualization of the function $x^x$ that I described in this answer.

It's not clear to me how to explain it to middle school kids, though. Specifically, if you want to explain the WolframAlpha output to middle schoolers, then they've got to know that
$$(-2)^x = e^{x\log(-2)} = e^{x(\log(2) + i\pi)} = 2^x e^{xi\pi} = 2^x(\cos(\pi x)+i\sin(\pi x).$$
Note that this last expression can be computed in terms of real trig and exponential functions. I'm guessing that most middle schoolers aren't quite there yet but, assuming your students are, then we can efficiently compute all kinds of values of your function. For example, what's $(-2)^{1/4}$? Well,
$$(-2)^{1/4} = \sqrt[4]{2}(\cos(\pi/4)+i\sin(\pi/4) = \sqrt[4]{2}\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = \frac{1}{\sqrt[4]{2}}+\frac{i}{\sqrt[4]{2}}.$$
We can do your $(-2)^{\pi}$, as well.
$$(-2)^{\pi} = 2^{\pi}(\cos(\pi^2)+i\sin(\pi^2)) \approx -7.96618 - 3.7974 i.$$

Now, if we can accept this, the Wolfram Alpha plot is then a simple matter of plotting the real part of $f(t)=(-2)^t$ together with the imaginary part, both on the same set of axes.

Note that the imaginary graph crosses the horizontal axis right at the integers and this accounts for some of the real values of $(-2)^x$. What about the other values? Well, it turns out that the complex logarithm has many branches corresponding to the fact that the exponential function is periodic in the complex direction. This can be accounted for in our formula for $(-2)^x$ by writing
$$(-2)^x = e^{x\log(-2)+2ki\pi} = e^{x(\log(2) + i\pi+2ki\pi)} = 2^x e^{i(2k+1)\pi x} = 2^x(\cos((2k+1)\pi x)+i\sin((2k+1)\pi x)).$$
Thus, we have a different plot for each $k$. If we plot these in sequence, we get something like so:

If we plot just the dots for a number of these, we get an image that is something like your blue graph.

---

As in my other answer, we can parametrically plot each curve $(t,\text{Re}(f_k(t)),\text{Im}(f_k(t)))$ in 3D. The result looks like so:

The thick path is the so-called principal branch, where $k=0$. The other paths are other branches. From this perspective, the points where $f(t)$ might be thought of as real arise as points where one of those paths punctures the $xy$-plane.