I borrowed the book Geometry of Complex Numbers by Schwerdtfeger. The material has been fascinating so far, though admittedly it took me about 3 hours to comprehend the material in page 1 of chapter 1. The first subject is this intriguing idea that you can represent a circle as a matrix. More specifically, you you can represent a circle with (complex) center 𝛾 and radius 𝜌 as a matrix

Where B = – A𝛾̅, C = -A𝛾, and D = A(𝛾𝛾̅-𝜌²), so that A and D are real numbers, and B and C are complex numbers. For normal circles, you will have A=1, though you can scale the matrix to have other (non-zero) values of A and still have the same circle. (If A is zero, you end up with a straight line, or some other mysterious thing that is not a circle.)

The simple case is the unit circle:

Which is

More exotic is the circle away from the origin, centered at 10+10i, i.e., (10, 10), with radius 3:

Here is some code to generate the matrix in Racket, and to put it back:

lang racket
(require math/array)
(require math/matrix)
(define (circle-to-matrix zC r)
(let ([B (* -1 (conjugate zC))]
[mzC (magnitude zC)])
(matrix [[ 1 B ]
[ (conjugate B) (- (* mzC mzC) (* r r)) ]])))
(define (matrix-circle-radius M)
(let ([A (array-ref M #[0 0])]
[d (matrix-determinant M)])
(sqrt (/ d (* -1 (* A A))))))
(define (matrix-circle-center M)
(let ([C (array-ref M #[1 0])]
[A (array-ref M #[0 0])])
(/ C (* -1 A))))

Something to explorer with the mandelbrot fractals is altering the n parameter in Z1 = Z^n. A slight tweak from the standard n=2 gives some interesting texture and pattern:

Something I would like to explore more is mandelbrots with an imaginary n (e.g., 3+2i) but my version of Fraqtive only supports real values of n.

One can generate a few slight variations on the appearance of the previous fractal by altering the scaling factoring of the circle radius a little:

It is interesting also that this fractal draws in such a way that you can paint alternate cells without painted cells touching each other: I.e., they only touch at the vertices.

This fractal idea did not originate with me, but I wrote some racket code to do the midpoint calculation using barycentric coordinates. This fractal draws a circle at the midpoint of a triangle, then subdivides the triangle and repeats:

Here is the same fractal to four iterations:

To get the midpoints, I could simple pass in the coordinates of the last triangle ABC, and then use “0.5” barycentric coordinates:

[P1 (barycentric->complex 0.5 0.5 0 A B C)]
[P2 (barycentric->complex 0.5 0.0 0.5 A B C)]
[P3 (barycentric->complex 0 0.5 0.5 A B C)]

An interesting concept I came across recently in an introductory geometry book is barycentric coordinates. The fundamental idea is defining the location of a point by its relationship to the points of a triangle. (Well, technically you can use any polygon, but we will keep things simple.)

So, each of the vertices of the triangle (A, B, and C) are given a sort of weight or pull factor. These three numbers uniquely represent the position of P. The theorem goes

If A, B, C are three non-collinear points, any vector P may be expressed in terms of the vectors A, B, C thus:

P = xA + yB +zC, where x + y + z = 1

Geometry: A Comprehensive Course, Dan Pedoe

If P is pulled all the way over to vertex A, then the coordinates become (1, 0, 0). Likewise at B it is (0, 1, 0) and for C it is (0, 0, 1). Another coordinate inside the triangle would be (⅓, ⅓, ⅓). If the point moves outside the triangle, some of those coordinates become negative.

Of course, instead of boring old vectors, you can use… complex numbers!

From the formula, it is obvious how to start with some barycentric coordinates, add a triangle, and find the point. Here is some Racket code:

(define (barycentric->complex x y z A B C)
(+ (* x A) (* y B) (* z C)))

Note, that you can use any triangle you want with the same barycentric coordinates. This could allow you to stretch the graph of some set of coordinates in various ways.

But, how do you convert in the other direction, having a point and a triangle, and wanting the barycentric coordinates? This is a little more complicated. Fortunately, it happens that the factor for each of the vertices is equal to the proportional area of the triangle opposite that vertex. So, looking at this diagram again

x = [BCP] / [ABC], y = [CAP] / [ABC], z = [ABP] / [ABC] where [BCP] is the area of triangle BCP, etc.

Now, a tricky part here is that the areas must be signed areas, or the formula doesn’t necessarily work correctly. Signed area means that you have to be consistent in assigning a negative or positive value to the area of the triangle, depending on the orientation of the vertices. So, if you happen to plug in the vertices in a counter-clockwise direction, you get a negative area, and in a clockwise direction, you get a positive area. Thankfully, there is an elegant formula for determining this, which I will represent by this Racket function

The function returns ‘ccw for counter-clockwise, ‘cw for clockwise, and ‘col for collinear (i.e., the points are on a straight line, which is not relevant in this application).

This allows us to have a signed-area function, using the common formula for determining the area of a triangle from just the length of the sides. We get the length of the sides by subtracting the complex numbers from each other and pulling the magnitudes of the resulting complex numbers.

(define (signed-area A B C)
(let* ([a (magnitude (- C B))]
[b (magnitude (- A C))]
[c (magnitude (- A B))]
[s (* (+ a b c) 0.5)]
[area (sqrt (* s (- s a) (- s b) (- s c)))]
[orient (orientation A B C)])
(if (eq? orient 'ccw)
(* -1 area)
area)))

And finally we determine x, y, and z:

(define (complex->barycentric A B C P)
(let ([ABC (signed-area A B C)])
(list (/ (signed-area B C P) ABC)
(/ (signed-area C A P) ABC)
(/ (signed-area A B P) ABC))))

In the previous post I was exulting about the coolness of solving a trigonometry problem without trigonometry, using complex numbers. Well, that isn’t quite the full truth. You can certainly do the problem described easily and elegantly with a complex number capable calculator. However, there is some hidden trigonometry in the initial rotation operation:

The complex number e^{i} has to be raised to a power. If you raise to an integer power, you could simply multiply out the rectangular values algebraically. But how do you raise a complex number to a fractional power, like 2.5? The standard approach, near as I can tell, is to convert the complex number to polar format, with a magnitude r and an angle ϕ. Assuming complex number a+bi, pulling out the magnitude requires only the sqrt function (pythagorean theorem) but pulling out the angle requires the arctangent of b/a. Then, use the formula Z^{n} = r^{n} (cos nϕ + i sin nϕ). I really don’t know if this is how all calculators do it, but here is the code behind the operations in the racket interpreter:

Scheme_Complex *cb = (Scheme_Complex *)base;
Scheme_Complex *ce = (Scheme_Complex *)exponent;
double a, b, c, d, bm, ba, nm, na, r1, r2;
int d_is_zero;
if ((ce->i == zero) && !SCHEME_FLOATP(ce->r)) {
if (SCHEME_INTP(ce->r) || SCHEME_BIGNUMP(ce->r))
return scheme_generic_integer_power(base, ce->r);
}
a = scheme_get_val_as_double(cb->r);
b = scheme_get_val_as_double(cb->i);
c = scheme_get_val_as_double(ce->r);
d = scheme_get_val_as_double(ce->i);
d_is_zero = (ce->i == zero);
bm = sqrt(a * a + b * b);
ba = atan2(b, a);
/* New mag & angle */
nm = scheme_double_expt(bm, c) * exp(-(ba * d));
if (d_is_zero) /* precision here can avoid NaNs */
na = ba * c;
else
na = log(bm) * d + ba * c;
r1 = nm * cos(na);
r2 = nm * sin(na);

The code there is just slightly more complicated, as it has separate angle finding conditional branches for real number and complex number exponents.

Of course, you don’t need to deal with any of that if you are just manipulating complex numbers in the interpreter or calculator.

I feel this would be a great place to dive into the interesting subject of complex roots, but that I should hold off until I have a more thorough grasp of the subject.

With the satellite dish pointed up 13 degrees above the ground, how far up will the Line of Sight (LOS) be after passing over 100 meters of ground? (To save a few sentences in our post, we will ignore the fact that the satellite dish will usually start a few meters above the ground.)

This is simple using any calculator with the trigonometry function tangent:

tan(ϕ) = opp/adj, so height = 100 * tan 13° (after 13° is converted to radians).

If you didn’t want to use a trigonometric function, however, what could you do? Another approach is complex numbers, and it is pretty simple. You simply need to rotate ϕ degrees up the unit circle (a basic operation with complex numbers), get the real and imaginary parts of that complex number, and use that proportion to figure out the opposite side of your problem triangle.

So, after raising e^i by 13 degrees (converted to radians) multiply 100 x (imag Z / real Z). In programming code, it looks like this:

What if you don’t have or want to use the exp function (to raise e to some value). You instead start with an approximation of e^i, which happens to be

0.5403023058681398+0.8414709848078965i

and then just raise that to ϕ.

Now, why would you want to use the complex number approach rather than the trig function? Well, to be honest I couldn’t think of a really compelling reason off the top of my head. An easy one would be if you were already using complex numbers for other reasons. Another would be if you wanted to use purely algebraic operations in your programming, say, on a computer board that did not have trigonometric functions built in. I suspect that the complex number approach is ultimately simpler and more efficient on the computational level, but since trigonometric functions are usually approximated with tables and with an algebraic exponential function, I couldn’t really say for sure off the top of my head.

In any case, it seems cool that complex numbers allow you to do an angle and distance trigonometry problem without trigonometry.