Ln of a complex number

[Albert Chan]

(03-03-2024 07:01 PM)johnb Wrote:  Albert, quite often, I really have to re-read your posts multiple times until my head hurts, or go dig out the math books and spend an enjoyable lunch break (or longer) before I really understand... but quite honestly, I've needed a friend/colleague like this for probably my entire career!

Thanks! I am not a mathematician, just learning as I go.

Solving a cubic is really solving a quadratic.
see https://www.hpmuseum.org/forum/thread-10...#pid150296

BTW, I don't like this form:
https://www.quantamagazine.org/the-scand...-20220630/


This may fail if we pick the wrong cube root (principle cube root may not work)
Also, it is inefficient, required evaluations of 2 different cube roots.
Also, it is inaccurate, 1 cube root may be hit with huge cancellation errors.

Thus, cubic_ab(a,b) only evaluate 1 cube root.
Now the issue of picking wrong cube root is gone.

cubic_ab(a,b)      → (α=(a/3)/β, β)      → αβ = a/3, guaranteed
x³ = (a*x+b)      → x = [α+β, α*ω+β/ω, α/ω+β*ω]

Cas> r := cubic_ab(15.,4.)      → [2-i, 2+i]
Cas> product(r)                      → 5.
Cas> sum(r)                           → 4.      // one root of cubic: x^3 = 15x + 4