Imaginary solutions outside the circle

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imaginary

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A circle (x^2+y^2)=r has imaginary (hyperbola) solutions for x or y > r. So do ellipses. But for ellipses, the imaginary hyperbolas are fixed to the orientation of the major and minor axes of the ellipse, and so are independent of the origin and orientation of the coordinate frame.

But a circle does not have major/minor axes; the orientation of the imaginary hyperbolas outside r are instead fixed to the choice of coordinate frame. Is this correct or am I missing something? It just seems counter-intuitive as the concept of a circle is something that is radially symmetric, and all of its properties should be choice of coordinates. If circles are a unique quadratic in which imaginary solutions depend on choice of reference frame, has this never caused a problem in math or physics?
 
Things should make sense.

Why doesn't a circle being a special case of an ellipse solve your dilemma?

[math]x^{2} + (y+\epsilon)^{2} = r^{2}[/math], where [math]|\epsilon| > 0[/math], shouldn't be TOO MUCH different from [math]x^{2} + y^{2} = r^{2}[/math]. No?

Where do these imaginary constructs have application?
 
Thanks for humoring my confusion.. Your example ellipse has imaginary hyperbolas along x and y axes, so smoothly transitions to the special case of the circle.

But this limit breaks if the ellipse is rotated eg 45 clockwise (or coordinates 45 degrees ccw). In that case the imaginary hyperbolas are 45 degrees off axis as ϵ approaches 0, but become "on axis" at ϵ = 0. So the imaginary solutions for a circle depend on subjective choice of coordinates ?.

I don't know of any application where this would matter, just a thought experiment. It is not analogous to the real and imaginary components of wave functions in quantum mechanics for example.

It just seems very weird that any, even imaginary property of a geometric object could possibly depend on subjective choice of coordinates. Imaginary components of functions in math and physics generally reflect some real thing, and are not simply artifacts of our subjective description. Also the circle seems to be the only function where this is the case. But I don't see this quirk mentioned anywhere, so maybe the weird is just in my thinking about it :unsure:
 
There is an alternative way of looking at conics. This is known as the polar equation for conics, it has nothing imaginary in it but maybe you will find it useful. It is very useful when dealing with central forces. 1595515715264.png
 
Yes the imaginary parts are not apparent in polar coordinates.

But I am specifically interested in the imaginary parts outside the ellipse. Particularly an ellipse with directrix at an angle to the x-axis (add some phase phi to theta in polar form). As eccentricity approaches zero, the imaginary hyperbolas flanking the vertices of the ellipse remain at the same angle phi off the x-axis, but suddenly align with the choice of coordinates when eccentricity = 0.
 
I am stuck on your first post.

[MATH]x^2 + y^2 = r^2[/MATH] has an infinite solution set in the reals.

If we graph that solution set in the cartesian plane, we get a circle with a center at the origin. You say that the circle has solutions. What does it mean to say that a solution set has solutions?

You then go to discuss the complex solution set, which respresents ordered pairs of complex numbers, which cannot be graphed in a cartesian plane. What do you mean that the solution is a hyperbola, which is a figure in a plane?
 
JeffM said:
I am stuck on your first post.

[MATH]x^2 + y^2 = r^2[/MATH] has an infinite solution set in the reals.

If we graph that solution set in the cartesian plane, we get a circle with a center at the origin. You say that the circle has solutions. What does it mean to say that a solution set has solutions?
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I have no idea what it means to say that a set has solutions and would never say such a thing!

You then go to discuss the complex solution set, which represents ordered pairs of complex numbers, which cannot be graphed in a cartesian plane. What do you mean that the solution is a hyperbola, which is a figure in a plane?
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Again, I cannot imagine such a thing. Where did you see this?
If you are still talking about [math]x^2+ y^2= r^2[/math] then I would say that, for any x in the complex numbers, there exist two values of y, [math]y = \pm\sqrt{r^2- x^2}[/math] such that (x, y) satisfies that equation.
 
Solve for y when |x| > r. Then y is imaginary, and takes the form of a hyperbola.

Same goes when solving for x when |y| > r.
 
imaginary said:
Solve for y when |x| > r. Then y is imaginary, and takes the form of a hyperbola.

Same goes when solving for x when |y| > r.
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When you say, |x| > r, you are restricting x and r to real numbers.

y is a number, not a hyperbola. Or if you prefer, y is a point in the complex plane. But possibly you are referring to the locus of points in the complex plane for y as x moves along the real number line. That is, we assume y is a complex number, and we graph y in the complex plane for different real values of x.

[MATH]x^2 + y^2 = r^2 \text { and } r > 0.[/MATH]
[MATH]|x| \le r \implies x^2 \le r^2 \implies \exists \text { real } a \ge 0 \text { such that } x^2 + a = r^2 \implies\\ r^2 - x^2 = a \ge 0 \implies y^2 = r^2 - x^2 = a \implies y = \pm \sqrt{a} + 0i.[/MATH]That locus of points is the horizontal axis (as the complex plane is usually oriented). In other words, it is a straight line, not a hyperbola.

[MATH]|x| > r \implies x^2 > r^2 \implies \exists \text { real } a < 0 \text { such that } x^2 + a = r^2 \implies\\ r^2 - x^2 = a \ge 0 \implies y^2 = r^2 - x^2 = a \implies y = 0 \pm i\sqrt{a}.[/MATH]That locus of points is the vertical axis (as the complex plane is usually oriented). In other words, it is a straight line, not a hyperbola.

So, demonstrate the plane in which this purported hyperbola exists and how it is derived.
 
I know what imaginary is doing, because I thought about the same graph when I was in high school. When I later learned about how complex variables are really worked with (letting both variables be complex), I understood that there wasn't much significance to it.

He (I'll assume) is graphing [MATH]x^2 + y^2 = 1[/MATH] allowing y to be a complex number, but keeping x real, by plotting y using the z axis for imaginary y. So for [MATH]|x|\le 1[/MATH], we have a circle in the xy plane, while for [MATH]|x|\ge 1[/MATH] we have a hyperbola in the x-z plane.

1595617669307.png

But the questions about rotational symmetry have no significance I can see, because by treating x and y so differently, any possible symmetry is lost.

The same graph can be made for an ellipse with an axis on the x axis, and will look the same; if the axis is rotated (in the x-y plane) the result would look quite different; I'll have to play with it to see how different. @imaginary, have you actually done that and seen the results? What equation did you use?
 
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Exactly Dr Peterson. I will try and make some figures... any advice on a free plotting tool?
 
I did this in GeoGebra's 3D mode, using parametric forms for four curves.

While I'm here, this is what I got for a 2-by-1 ellipse rotated 45 degrees, namely 5x² - 6x y + 5y² = 8:

1595631842351.png
 
HallsofIvy said:
And I did not say that you did! I only said that I would not say such a thing!


And you have been told that it doesn't mean anything.
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OK. I interpreted your post differently. Sorry about that.

Now that I have seen Dr. P's response, I see what is behind this. But I do not see that focusing exclusively on one cross-section can tell us anything. In any case, it is the OP's responsibility to make clear what is being talked about. A hyperbola exists in a plane. If Dr. P is correct about what the OP is saying, (and I suspect he is), we are being asked to think about the locus of points in two different constrained planes. It would have been nice to have been told this up front.
 
Sorry for generating so much confusion, but the attention is much appreciated!

When I plot a circle and solve for y where |x|>r, the solutions are imaginary of course, and take the form of a hyperbola in the x-iy plane. Similarly, solving for x where |y|>r gives the other "imaginary hyperbola" in the ix-y plane.

Doing the same for an ellipse and plotting the "imaginary hyperbolas" in red:
1595645359777.png
I ASSUMED that a rotated ellipse would give:
1595645395605.png
...but this ASSUMPTION was WRONG.

I worked out a real example using 1595645464693.png, and solving for y along x I found:
1595645513528.png
The “imaginary hyperbolas” do NOT rotate along with the directrix, but rather remain on axis. Solving for y when x is outside the rotated ellipse (i.e. for |x|>sqrt(10/64)) gives complex, NOT purely imaginary solutions. The parabolas remain on-axis but the solutions instead have a linear (-6x/10 here) real component (only applies OUTSIDE the ellipse) whose slope is the line from origin to the perpendicular tangents (at x=+/-sqrt(6/10) here).

So now I see there is a smooth transition of these hyperbolas when varying eccentricity and rotation around 0. But that means that the imaginary hyperbolas on ellipses are fixed to the orientation of the coordinate frame we choose, not to the orientation of the ellipse.

That is even weirder than my original confusion! The ellipse is invariant but the hyperbolas are fluid to our choice of coordinates(?)

p.s. @Dr. P: I will try GeoGebra, thanks! The plots above are from FooPlot.
 
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Something you said reminded me of an error in my last image; here is what I think is a corrected version, which agrees with what you show (once I figured out what it meant), but not with your description, as they are indeed rotated, just not where you thought:

1595647416916.png

Here is a different view that shows the relationship better:

1595647518976.png

I think the main lesson may be, never assume you know what will happen!
 
Dr. P thanks again! Your picture is much clearer than my plotting real and imaginary parts separately. You are right the hyperbolas rotate.. a bit.

But they do not remain at the vertex of the ellipse. Rather, they extend along the line that intersects where tangents (|| to y) touch the ellipse.

The math speaks for itself, but the point remains that the hyperbolas move with respect to the ellipse if we rotate coordinates. It is intuitively very weird because the ellipse and the hyperbolas are THE SAME equation. I would think the shape of an equation should be invariant under rotation of coordinates.
 
...But the hyperbolas are really generated by projecting the equation onto y along x, and onto x along y. It makes more sense that these projections would not be invariant under rotation. What is invariant is the projection onto any axis along the orthogonal axis, as long as these axes are at the same angle to the major and minor axes. @Dr.Peterson is this what you meant above by: "by treating x and y so differently, any possible symmetry is lost"?
 
imaginary said:
Dr. P thanks again! Your picture is much clearer than my plotting real and imaginary parts separately. You are right the hyperbolas rotate.. a bit.

But they do not remain at the vertex of the ellipse. Rather, they extend along the line that intersects where tangents (|| to y) touch the ellipse.

The math speaks for itself, but the point remains that the hyperbolas move with respect to the ellipse if we rotate coordinates. It is intuitively very weird because the ellipse and the hyperbolas are THE SAME equation. I would think the shape of an equation should be invariant under rotation of coordinates.
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Can you see why the vertex of the hyperbola CAN'T be at the vertex of the ellipse? Hint: look at where they actually are, with respect to x.

and, no the ellipse and the hyperbola are not the same equation; they are only derived from the same equation.

imaginary said:
...But the hyperbolas are really generated by projecting the equation onto y along x, and onto x along y. It makes more sense that these projections would not be invariant under rotation. What is invariant is the projection onto any axis along the orthogonal axis, as long as these axes are at the same angle to the major and minor axes. @Dr.Peterson is this what you meant above by: "by treating x and y so differently, any possible symmetry is lost"?
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Yes, that's part of it. The main idea is simply that real x and complex y are not the same. The rotation in the real-x/real-y plane doesn't translate into the imaginary part. There's no reason to think it should.
 
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