Answer:
(6^⅕) (cos(-24°) + i sin(-24°))
Step-by-step explanation:
First, we convert from Cartesian to polar:
r = √((-3)² + (-3√3)²)
r = √(9 + 27)
r = 6
θ = atan( (-3√3) / (-3) ), θ in the third quadrant
θ = atan(√3)
θ = 240° + 360° k
Notice that θ can be 240°, 600°, 960°, etc.
Therefore:
-3 − 3√3 i = 6 (cos(240° + 360° k) + i sin(240° + 360° k))
Now we take the fifth root:
[ 6 (cos(240° + 360° k) + i sin(240° + 360° k)) ]^⅕
(6^⅕) [ (cos(240° + 360° k) + i sin(240° + 360° k)) ]^⅕
Applying de Moivre's Theorem:
(6^⅕) (cos(⅕ × 240° + ⅕ × 360° k) + i sin(⅕ × 240° + ⅕ × 360° k))
(6^⅕) (cos(48° + 72° k) + i sin(48° + 72° k))
If we choose k = -1:
(6^⅕) (cos(-24°) + i sin(-24°))
