Ln of a complex number

[Albert Chan]

(03-03-2024 08:23 PM)Albert Chan Wrote:  Solving a cubic is really solving a quadratic.
see https://www.hpmuseum.org/forum/thread-10...#pid150296

Perhaps it is easier to understand if we matched cubic_ab(a,b) code.
Starting from identity, with ω = cis(2*pi/3)

x³ - 3αβx - (α³+β³) = (x-(α+β)) * (x-(αω+β/ω)) * (x-(α/ω+βω))

LHS is a depressed cubic: x³ - a*x - b , where a = 3αβ, b = (α³+β³)

We setup a quadratic of t, with roots (α³, β³)

(t - α³)*(t - β³) = t² - b*t + (a/3)³ = 0
t = b/2 ± √Δ, where Δ = (b/2)² - (a/3)³

β = ³√t
α = (a/3) / β

x³ = a*x + b    → x = [(α+β), (αω+β/ω), (α/ω+βω)]