Hyperbolic numbers are of the form a + bj, where jÂČ = 1, but j â ±1, and a, b â â. Adding them works the same with complex numbers: you add their real and hyperbolic parts together. Multiplying them works a bit differently. For example, (a + bj)(c + dj) = (ac + bd) + (bc + ad)j whereas in the complex numbers, (a + bi)(c + di) = (ac - bd) + (bc + ad)i.
In complex numbers, solving quadratic equations of the form az2 + bz + c = 0 is straightforward. If a â 0, the solutions are z0 = (-b + â(b2 - 4ac))/2a, and z1 = (-b - â(b2 - 4ac))/2a. In hyperbolic numbers, a quadratic equation can have up to 4 roots, not just 2, when a, b, c, z â đ» (the hyperbolic numbers). Any hyperbolic number can be mapped into an isomorphic direct sum of two real numbers using the orthogonal idempotent basis:
u = (1+j)/2, v = (1-j)/2
these idempotent elements satisfy: u2 = u, v2 = v, uv = 0, and u+v = 1.
then, we express a, b, c, and z in terms of their idempotent components: a = a_1 u + a_2 v, b = b_1 u + b_2 v, c = c_1 u + c_2 v, and z = z_1 u + z_2 v. (where a_1 = a_real + a_hyper, a_2 = a_real - a_hyper, b_1 = b_real + b_hyper, b_2 = b_real - b_hyper, c_1 = c_real + c_hyper, c_2 = c_real - c_hyper, and z_1 = z_real + z_hyper, z_2 = z_real - z_hyper)
Next, we decouple it into two quadratic expressions with real components, for z = z_1 u + z_2 v:
a_1 z2_1 + b_1 z_1 + c_1 = 0
a_2 z2_2 + b_2 z_2 + c_2 = 0
These can be solved using the quadratic formula:
z_1 = (-b_1 ± â(b2_1 - 4 a_1 c_1))/2a_1
z_2 = (-b_2 ± â(b2_2 - 4 a_2 c_2))/2a_2
This gives 4 roots. Substituting them back to find z gives:
z = (z_1 + z_2)/2 + j(z_1 - z_2)/2
Likewise, a cubic equation in đ» can have a maximum of 9 solutions, a quartic 16, and generally, an n degree polynomial can have up to n2 solutions in đ».
But what if the hyperbolic number z is in the exponent, not the base? Here's how such equations can be solved.
From the Taylor series, we know:
exj = 1 + xj + x2j2/2! + x3j3/3! + x4j4/4! + ...
Which equals:
1 + xj + x2/2! + x3j/3! + x4/4! + ... = (1 + x2/2! + x4/4! + ...) + (x + x3/3! + x5/5! + ...)j
The real and hyperbolic parts are the Taylor expansions for cosh(x) and sinh(x) respectively. This gives us a hyperbolic version of Euler's formula, where instead of sine and cosine, we have the hyperbolic versions of these functions:
exj = cosh(x) + sinh(x)j
We can use this to solve a simple exponential equation of the form az = b, with a â â+. We start by converting az into ax+yj (x, y â â) and by exponent rules, this equals ax * ayj. a = eln(a), so ayj = e(y ln(a)j) = cosh(y ln (a)) + j sinh(y ln (a)). Therefore, az = ax cosh(y ln (a)) + j ax sinh(y ln (a)) = b. This gives a system:
ax cosh(y ln (a)) = b_real
ax sinh(y ln (a)) = b_hyper
Solving this system isn't easy, but the problem is now reduced to â.
If you have any questions, suggestions, or are interested to learn more about this system, comment below.