Express your answer in the form found using Euler's Formula, The cube roots of 2 + 3i

Answer 1

let z = 2+3i = √13(cos 56.3099..° + i sin 56.309.°) or √13cis 56.3099°
primary cube root
= z^(1/3) = (√13)^(1/3)cis 18.7699..° = (√13)^(1/3) e^(i 18.7699..°)

two more cube roots would be found by adding 360/3° or 120°

which would be (√13)^(1/3) e^(i 138.7699..)
and (√13)^(1/3) e^(i 256.7699...)

btw, another form of (√13)^(1/3) e^(i 18.7699..°) = 1.4519 + i(.4934)

I found this using degrees, I you want radians your result would be
(√13)^(1/3) e^(i .3276) , add 2π/3 to the argument to get the other two.