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General Science => General Science => Topic started by: chiralSPO on 11/02/2018 21:54:21

Title: what are the complex solutions for x[sup]2[/sup] + y[sup]2[/sup] = 1 ?
Post by: chiralSPO on 11/02/2018 21:54:21
The familiar equation for the unit circle centered at the origin of the (x,y) coordinate system is:
 x2 + y2 = 1

Using only real numbers for both x and y only yields useful solutions for 1 ≤ x ≤ 1 and 1 ≤ y ≤ 1. But if we allow for either x or y to have nonzero imaginary components (x or y can be imaginary or complex), then the solution is no longer just a unit circle, but an odd 2-dimensional surface in 4-dimensional space (x, a, y, b) such that (x + ai)2 + (y + bi)2 = 1.

Try as I might, I am having great difficulties trying to visualize this surface. I can generate a few different 3-d slices, like a = 0 for which all of the points have only real components of the x coordinate. This slice contains a unit circle in the x,y plane, and a hyperbola in the x,b plane (x2 b2 = 1). Similarly for the b = 0 slice we also see the same unit circle and a hyperbola in the y,a plane (y2 a2 = 1).

Which other slices would be useful for me to try to graph to visualize the rest of this shape? (I am tempted to look at x=y or just something like several parallel planes using integer values of a or b.
Title: Re: what are the complex solutions for x[sup]2[/sup] + y[sup]2[/sup] = 1 ?
Post by: chiralSPO on 11/02/2018 21:54:37
suggestions? is there an obvious name for this shape?
Title: Re: what are the complex solutions for x[sup]2[/sup] + y[sup]2[/sup] = 1 ?
Post by: jeffreyH on 11/02/2018 22:54:36
The function for the unit circle is a difficult one. It is a bit late here, just going to bed, but I will give this some thought tomorrow.
Title: Re: what are the complex solutions for x[sup]2[/sup] + y[sup]2[/sup] = 1 ?
Post by: evan_au on 12/02/2018 07:47:18
I thought that there was a principle in mathematics that a quadratic equation (one with x2) had 2 zeros (or roots), some of which might be imaginary.

For x2 + y2 = 1, there are two real roots, at (-1,0) and (+1, 0).

But I'm not sure how this applies to a formula where there is a y2 as well... :-\
Title: Re: what are the complex solutions for x[sup]2[/sup] + y[sup]2[/sup] = 1 ?
Post by: syhprum on 12/02/2018 10:04:21
I believe Mathematica  shows plots of such functions free trial versions are available.
Title: Re: what are the complex solutions for x[sup]2[/sup] + y[sup]2[/sup] = 1 ?
Post by: evan_au on 12/02/2018 10:18:45
Thinking about it some more...

x2 + y2 = 1 can also be written as: y = SQRT(1 - x2)

The part inside the square root is a parabola z = 1 - x2 = (1-x)(1+x)
...which has two zeroes, at x = +1 and x= -1.

When you take the square root of a number, SQRT(0) = (0), so these two zeroes remain.
- When you take the square root of a positive number, the answer is real (both positive and negative), so no new zeroes appear.
- When you take the square root of a negative number, the answer is imaginary (both positive and negative), so no new zeroes appear.
- When you take the square root of an imaginary number (apart from the origin): I have a vague recollection that you can work it out in polar coordinates: The radius changes (but not to 0), and there is a rotation about the origin. If correct, this would mean that no new zeroes appear.

So I suggest that this equation has just two zeroes, for all values of x in the imaginary plane. In the case givem both zeroes occur for real x.

PS: I heard about an ancient Persian mathematician who worked out how to solve quadratic equations.
If I recall correctly, he had to represent 6 classes of quadratic equations, because he couldn't grasp the notion that "something" could equal "nothing".
While today we often represent the solution of all these quadratic equations as "=0".
See: http://www.bbc.co.uk/programmes/b006qykl/topics/Medieval_Persian_mathematicians
Title: Re: what are the complex solutions for x[sup]2[/sup] + y[sup]2[/sup] = 1 ?
Post by: jeffreyH on 12/02/2018 12:36:43
The equations are: y = SQRT(1 - x^2), -iy = SQRT(x^2 - 1), x = SQRT(1 - y^2) and -ix = SQRT(y^2 - 1). I may be wrong. Please correct if so.
Title: Re: what are the complex solutions for x[sup]2[/sup] + y[sup]2[/sup] = 1 ?
Post by: evan_au on 12/02/2018 20:26:55
Quote from: evan_au
When you take the square root of an imaginary number
I worked it out overnight...
If you take a number on the complex plane, you can represent it in polar coordinates as (r, θ).

The two square roots of (r,θ) lie on a circle of radius SQRT(r), centered on the origin.
One root lies on the circle at angle θ/2, and the other lies 180 around the circle.

And the only way the square root of a number can be zero is if the original number has r=0 (ie the origin).
So the square root function does not produce any more zeroes in the formula of a circle, when calculated on the complex plane.

...but when I looked it up in Wikipedia, this method was already described (well, in the meantime it gave a few grey cells a workout :) )
See: https://en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number