e^ifi=i+1 then 8i+8=? = 1/e^fi ??? question is then: 1/e^fi is when 8i=>i^8=-1? then e^fi=0.125=1/8; e^x=1/8 => x=fi=>x=2.08 and 2.72^2.08=8 I'm not mathematician, just story teller 😎😉😃 The check: w=8i; then 8i=8+8i => i^8=8(i+1) => -1=8(i--1) => -1=8i--8 => -1=-1-i^8 => -1=0 ; when i^2=0?? if cos(fi)=npi and 00cos(npi)=i * i * cos(npi) =>0 * 0 * cos(90)=i * i * cos(90)

All right enough of stories! How could we go for this in real brute force attack? THIS IS MY REAL IDEA:

using Pkg
Pkg.add("Plots")
using Plots

# Tvoje hodnoty k (predstavujú imaginárnu časť: 8 + k*im)
ks = [
    4, 3, 1.57, 10, 13, 15, 19, 23, 33, 43, 53, 63, 73, 83, 100, 120, 130,
    140, 150, 180, 200, 220, 320, 360, 380, 450, 500, 550, 555, 565, 595,
    695, 795, 1000, 1200, 1300, 1500, 1800, 2800, 3600, 4600, 5600, 5720
]

# Automaticky vypočítané pomery angle(8+kim)/angle(kim)
vals = [ angle(8 + k*im) / angle(k*im) for k in ks ]

# Histogram
scatter(vals,
    bins = 20,
    ylabel = "Ratio angle(8+kim) / angle(kim)",
    xlabel = "K multip",
    title = "Discrete graph: solution w=8+w; w=5720*im",
    legend = false
)
display(current())
# Pre interaktívne prostredie
if !Base.isinteractive()
    println("Press enter to quit:")
    readline()
end

What I was looking for was this trick:

Kód vypľula AI gpt, fakt neovládam naspamäť Julia lang, použil som svoj nápad pre prompt a kód je automatizovaný, mierne upravený mnou ^^^

---