Question

Find the complex cube roots of 8(cos(4pi/5)+ isin(4pi/5)).

Answer 1

We have to find the cube roots of complex no 8[cos(4π/5) + i sin(4π/5)]

solution : complex no is x = 8[cos(4π/5) + i 8sin(4π/5)]

x⅓ = 8⅓ [cos(4π/5) + i sin(4π/5)]⅓

from DeMoivre’s theorem,

x⅓ = y = 2[cos(2πn + 4π/15) + isin(2πn + 4π/15) ]

where n = 0, 1 , 2 ...

Therefore the cube roots of given complex number is 2[cos(2πn + 4π/15) + isin(2πn + 4π/15)]

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