Find the fifth roots of 32(cos 280° + i sin 280°).

Answer 1

Let, z = 32(cos 280⁰ + i sin 280⁰) = 32{ cos(360⁰*n +  280⁰) + i sin(360⁰*n +  280⁰) }, where n is an integer.

⇒ z^1/5 = [ 32{ cos(360⁰*n +  280⁰) + i sin(360⁰*n +  280⁰) } ]^1/5

From De - Moievere's theorem.

z^1/5 = 2{ [cos(360⁰*n +  280⁰)/5] + [i sin(360⁰*n +  280⁰)/5] }

⇒ z^1/5 = 2{ cos(72n + 56) + i sin(72n + 56) }, where n is an integer.