_{0}= 2(cos(4pi/15)+isin(4pi/15)) = 1.33826 + 1.48629i

_{1 = 2(cos(14pi/15)+isin(14pi/15)) = -1.95623 + 0.41582i}

_{2 = 2(cos(24pi/15)+isin(24pi/15)) = 0.61803 - 1.90211i}

I just need help. If someone could give me the answer as well as explain it, that would be amazing

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If z is a complex number, written in polar form as

z=8(cos(4pi/5) + isin(4pi/5)

then z^(1/3) = (8(cos(4pi/5) + isin(4pi/5))^(1/3)

= 8^(1/3)(cos((1/3)(4pi/5+2pi*k)) + isin((1/3)(4pi/5+2pi*k))) where k=0,1,2

when K = 0, arg z= 4pi/15 the root a_{0} = 2(cos(4pi/15)+isin(4pi/15)) = 1.33826 + 1.48629i

k = 1, arg z= 14pi/15 a_{1 = 2(cos(14pi/15)+isin(14pi/15)) = -1.95623 + 0.41582i}

k = 2, arg z= 24pi/15 a_{2 = 2(cos(24pi/15)+isin(24pi/15)) = 0.61803 - 1.90211i}

DeMoivre's Theorem tells that there are always three cube roots of a complex number, spaced evenly around a circle.

if w=r(cosΘ+isinΘ), then z_{k}=cube root(r)•(cos((Θ+2πk)/3)+isin((Θ+2πk)/3)), k=0,1,2.

Please google "roots of complex numbers" to find more information on the topic with examples and illustrations.

Good luck!

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