Find the fifth roots of 243(cos 260° + i sin 260°) 

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Use de moivre's theorem: ⁵√[243(cos 260° + i sin 260°)] = [243(cos 260° + i sin 260°)]^(1/5)= 243^(1/5) (cos (260 / 5)° + i sin (260 / 5)°)= 3 (cos 52° + i sin 52°)z1 = 3 (cos 52° + i sin 52°) ←← so that's the first rootthere are 5 roots so the angle between each root is 360/5 = 72°then the other four roots are: z2 = 3 (cos (52 + 72)° + i sin (52+ 72)°) = 3 (cos 124° + i sin 124°)z3 = 3 (cos (124 + 72)° + i sin (124 + 72)°) = 3 (cos 196° + i sin 196°)z4 = 3 (cos (196 + 72)² + i sin (196 + 72)°) = 3 (cos 268° + i sin 268°)z5 = 3 (cos (268 + 72)° + i sin (268 + 72)°) = 3 (cos 340° + i sin 340°)

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