Asked by Nick

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

Answers

Answered by Reiny
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..° = <b>(√13)^(1/3) e^(i 18.7699..°)</b>

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

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

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)</b> , add 2π/3 to the argument to get the other two.

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