Recall that:
-20i = 20(cos 3π/2 + i sin 3π/2)
Then, by De Movire's Theorem:
(-20i)^(1/4) = 20^(1/4) * [cos (3π/2 + 2πk)/4 + i sin (3π/2 + 2πk)/4]
==> (-20i)^(1/4) = 20^(1/4) * [cos (3π + 4πk)/8 + i sin (3π + 4πk)/8]
For k = 0, 1, 2, and 3.
For k = 0:
(-20i)^(1/4) = 20^(1/4) * (cos 3π/8 + i...