Question

Find each principal root. Express the result in the form a+bi with a and b rounded to the nearest hundredth.
(32(cos2pi/3 +isin2pi/3))^1/5

Answer 1

I saw you post it twice, what don't you understand?

Answer 2

oh I don't understand the last part of the answer

Answer 3

What is De Moivre's theorem?

Answer 4

well I got up to 2(cos2pi/15 +isin2pi/15)

Answer 5

\(if, z=r*(a *cos x + i*bsinx)\)
then \(z^{\color{red}{n}}= r^{\color{red}{n}}*(a cos(\color{red}{n}*x+i*b*sin(\color{red}{n}*x))\)
Your n is 1/5,
r = 32
a=1
b=1
x=2pi/3, just plug them all in
\(z^{\color{red}{1/5}}= 32^{\color{red}{1/5}}* cos(\color{red}{1/5}*(2pi/3)+i*sin(\color{red}{1/5}*(2pi/3))\)

Answer 6

\(32^{\color{red}{1/5}}= \sqrt[\color{red}{5}]{32}=2\)

Answer 7

\(\color{red}{1/5}*2pi/3= 2pi/15\)