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.
  
    
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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