Find the cube roots of 8(cos 216° + i sin 216°).

Find the cube roots of 8(cos 216° + i sin 216°).

Answers

z^3=8(cos216^circ+isin216^circ)
z^3=2^3(cos(6^3)^circ+isin(6^3)^circ)
implies z=8^{1/3}left(cosleft(dfrac{216+360k}3right)^circ+isinleft(dfrac{216+360k}3right)^circright)

where k=0,1,2. So the third roots are

z=begin{cases}2(cos72^circ+isin72^circ)\2(cos192^circ+isin192^circ)\2(cos312^circ+isin312^circ)end{cases}

Start at k = 0, 1, 2, because there are THREE cube roots of a complex number: 

k = 0: 2(cos 72° + i sin 72°) 
k = 1: 2(cos 192° + i sin 192°) 
k = 2: 2(cos 312° + i sin 312°)

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