# Between line and parabola

Take a look at the first three powers of $x$. Can you imagine how to smoothly go from one to another?

Line, parabola, and cubica

For positive $x$, the intuition that one can obtain higher powers of $x$ by simply “bending” and “stretching” the line $y = x$ seems satisfactory; moreover, lower powers (e.g., $\sqrt{x}$) can also be obtained in the same manner, which is reassuring. What about negative $x$?

The negative $x$’s are just a mess. If you consider going over all positive powers, the negative part of the graph will oscillate between positive and negative values forever! And how the heck is it going to transition between them? For example, when going from $x$ to $x^2$, the left tail of the graph has to cross the $x$-axis somehow. That seems bizarre, to say the least.

The solution is, of course, to consider the function $w = z^\alpha$ in the complex plane. We have to resort to complex numbers out of necessity: there is simply no other way to make sense out of expressions like $(-1)^\frac{3}{2}$.

## Natural powers


Consider $n = 2$ and point $z = 2\me^{i\frac{\pi}{2}} = 2i$. Function $w$ maps it to $w(z) = 4\me^{i\pi} = -4$. So, a point on the imaginary line goes into a point on the real line. In general, all points move from one line to another, except for some lines that preserve all their points—these lines are invariant subspaces of the function $w = z^n$.

Let’s find all lines that remain fixed under the transformation $w = z^n$. Points on a ray after transformation have to lie either on the same ray or on the ray rotated by $k\pi$

For $n = 2$, there is only one invariant line $\phi = 0$. It is understood that $\phi = 0$ and $\phi = \pi$ define the same lie. Within that line, the ray $\phi = 0, r \in [0, \infty)$ is itself an invariant subspace, whereas the ray $\phi = \pi, r \in [0, \infty)$ is not—all its points end up on the positive ray. However, the line as a whole is invariant. For $n = 3$, we obtain $\phi = \left\{0, \frac{\pi}{2}\right\}$, and for $n = 4$, $\phi = \left\{0, \frac{\pi}{3}, \frac{2\pi}{3}\right\}$.

## Rational powers

Considering integer powers does not intoduce anything new because $z^{-n} = (1/z)^n$, therefore we go straight to rational powers. Let’s first look at the simplest case $w = z^\frac{1}{2} = \sqrt{z}$.

The first remarkable observation is that the square root is a one-to-many function on $\bC$: for every $z$, it returns a set of complex numbers $w = \left\{s_1, s_2\right\}$, such that for every $s \in w$, $s^2 = z$. In general, $\sqrt[n]{z}$ returns $n$ numbers. One can make $\sqrt[n]{z}$ into a one-to-one function by cutting the plane and considering the function on its Riemann surface. To avoid such complications, we will restrict our attention to the principal branch of a given function, as it suffices to get the basic intuition. The principal branch $g_0$ of a function $\sqrt[n]{z}$ is the one-to-one function

from the complex plane with a cut from $-\infty$ to $0$ to the angular area $G_{-\frac{\pi}{n}, \frac{\pi}{n}} = \left\{ z \neq 0 \mid -\frac{\pi}{n} < \arg z < \frac{\pi}{n} \right\}$. So, $z^n$ is unfolding an angular area onto $\bC$ with a cut, whereas $z^\frac{1}{n}$ is folding $\bC$ with a cut onto an angular area. Rational powers are compositions of foldings and unfoldings, for $z^\frac{k}{m} = (z^\frac{1}{m})^k$.

## $w = z^\alpha$ on an arc

Now we can generalize $w = z^\alpha$ to all real $\alpha$. We can view function $w$ as a function of three real arguments $w = w(r, \phi, \alpha)$. It is interesting to see how it acts on subspaces $r = \mathrm{const}$ and $\phi = \mathrm{const}$, i.e., on arcs and rays. In this section, we will look at arcs.

Let’s fix $r = 4$, take several $\phi$’s from $[-\frac{\pi}{2}, \frac{\pi}{2}]$, for example, $\phi \in \left\{ -\frac{\pi}{2}, -\frac{3\pi}{8}, -\frac{\pi}{4}, -\frac{\pi}{8}, 0, \frac{\pi}{8}, \frac{\pi}{4}, \frac{3\pi}{8}, \frac{\pi}{2} \right\}$, and look at all $\alpha \in [-2, 2]$.

$z^\alpha$ for powers $\alpha \in [-2, 2]$ on the arc $r = 4, \phi \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

Points on the arc $r = 4, \phi \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ remain where they are for $\alpha = 1$. When $\alpha$ grows from $1$ to $2$, the right half-plane unfolds to the full plane with a cut $(-\infty, 0]$. When $\alpha$ goes from $1$ to $0$, all points shrink to a single point $1$. That point is a sort of singularity, akin to the singularity of the Big Bang. When $\alpha$ goes from $0$ to $-2$, the points unfold again, getting smaller in amplitude. It is interesting to note that arcs change orientation after passing through $1$. By the way, we’ve simultaneously described two arcs: $r = 4$ and $r = 1/4$.

## $w = z^\alpha$ on a ray

Let’s consider what happens to the ray $\phi = \pi/2$. We take $r \in \left\{ \frac{1}{3}, \frac{1}{2}, 1, 2, 3 \right\}$ and $\alpha \in [-2, 2]$.

$z^\alpha$ for powers $\alpha \in [-2, 2]$ on the ray $\phi = \frac{\pi}{2}$ for $r \in [\frac{1}{3}, 3]$.

Rays turn into rays for different $\alpha$; points on the unit circle remain on the unit circle. Otherwise, points outside the unit circle move inside as $\alpha$ goes from $2$ to $-2$, and those inside go outside in the opposite direction. All points meet at $1$. By the way, we’ve simultaneously described two rays: $\phi = \frac{\pi}{2}$ and $\phi = -\frac{\pi}{2}$.

## Conclusion

There is life between line and parabola, but it is complex. Here is the Jupiter notebook to generate the images in this post, Between line and parabola.