The image of a conic section under the $z^2$ map

Question

My question in short: In some cases, the image of a conic section under the $z^2$ map is still a conic section. Is there an elegant argument to show that?

Let $\Gamma$ be a conic section in the xy-plane. Consider the map $(x,y)\mapsto (x^2-y^2,2xy)$. What is the image of $\Gamma$ under the map? The five cases I'm interested in are:

Case 1: If $\Gamma$ is a line that passes the origin. Then the image is a ray starting at the origin.

Case 2: If $\Gamma$ is a line that does not pass the origin. Then the image is a parabola.

Case 3: If $\Gamma$ is a circle or ellipse centered at the origin. Then the image is a circle or ellipse.

Case 4: If $\Gamma$ is a hyperbola centered at the origin. Then the image is a line or a hyperbola.

Case 5: If $\Gamma$ is a parabola, the image is not necessary a conic section.

I basically verify this case by case. (With the help of the identity $(ze^{i\theta})^2=z^2e^{i\theta}$, we may rotate $\Gamma$ so that it is symmetric with respect to either the x-axis or y-axis.) Are my results correct? Is there more elegant way (to deal with Case 1 to 4 all at once) to get the same conclusion?

Thank you.

Answer 1

Are my results correct? Is there more elegant way (to deal with Case 1 to 4 all at once) to get the same conclusion?

This is not an answer, rather a suggestion for further study, written in the space of an answer because comment space is too short and is only an answer for your quoted question, above.

The complex map $w=z^2$ is not trivial enough on the $w$ plane to allow you to extract general conclusions so fast. Keep in mind that the same map iterated continuously, generates extreme complexity on the $w$-plane (Mandelbrot Set).

What the map does however, is fairly easily visible. Look at this after you type $z^2$ in the bottom.

The displayed $w$ plane should ideally answer all your questions about $\mathit{any}$ curve in question, by tracking a rough itenerary of any trajectory on the tiles.

If what this mapping does is not obvious enough on that webpage for you to answer all your questions above, then you need to break down your analysis to a quadrant by quadrant basis. It is obvious for example, that (for $w=z^2$) at least one angular contraction ($-\pi/2\le\theta\le\pi/2)\to(0\le\theta^*\le\pi/2)$ occurs in the first quadrant, paving the way for what looks like hyperbolic components.

In other words, some right half plane gets contracted into the first quadrant and as a result the map displays at least one hyperbolic "pintching" in that quadrant.

Your first due here then, would be to examine the images of the curves in question, inside and outside the unit circle in the first quadrant. Cases 1 and 4 are, I think, fairly obvious and here you can, of course, use some graphing software to draw a parametrized version of a particular curve $\gamma=f(x,y)$ you are interested in and then the curve $\gamma^*=f^{*}(x,y)=(x^2-y^2,2xy)$ and then you can compare what the transformation does for your particular case in that quadrant.

Generalizing, you will find that the map $w=z^n$, forces a right half plane to suffer at least one angular contraction of $\pi/n$ in the first quadrant ($-\pi/2\le\theta\le\pi/2)\to(0\le\theta^*\le\pi/n)$, paving the way for at least one hyperbolic component in that region.

As far as the answer to the general question goes, I would haphazard $\mathit{very}$ cautiously, that in general the final shape of any general Jordan curve under the map, wouldn't necessarily always give a conic section, since if I recall correctly in the Mandelbrot set, besides hyperbolic and parabolic components in Julia Sets, one may also get various other curves, such as spirals, from weak star limits in Siegel disks, etc.

But let me note that the above answer is just a guess and not a proof.