I) Let z = 32 (cos 280 degrees + i sin 280 degrees) = 32 (cos (k* 360 + 280) + isin (k*360 + 280)), where k = integer II) z^ (1/5) = (32 (cos (k*360 + 280) + i sin (k*360 + 280))) ^ (1/5) then, z^ (1/5) = 2(cos ((k*360 + 280)/5) + i sin((k*360 + 280)/5)) We apply De Moivre's theorem: z^ (1/5) = 2(cos(72k + 56) + i sin (72k +56)) We can get the five roots by assigning k = 0, 1, 2, 3, and 4 When K = 0, 1st root = 2 + i sin (cos 56 + i sin 56)k = 1, 2nd root = 2 (cos 128 + i sin 128)k = 2, 3rd root = 2 (cos 200 + i sin 200)k = 3, 4th root = 2 (cos 272 + i sin 272)k = 4, 5th root = 2 (cos 344 + i sin 344)
The fifth root is 5th root = 2 (cos 344 + i sin 344)
