find the cube root of 4 - 4 times the square root of 3i that graphs in the second quadrant.

Question

I need help trying to find the answer in terms of sine and cosine using degree measure.

Answer 1

First, express √i in terms of cos θ and sin θ.
De Moivre’s Theorem: e^i π/2 = cos π/2 + i sin π/2 = i
[e^iπ/2]1/2 = i ½ = cos π/4 + i sin π/4 = 0.707 + 0.707 i
Let r = 4 - 4√(3i) = 4 – 4 √3. √i = 4 – 4 √3 (0.707 + 0.707 i)
= -0.898 – 4.898 i
R = √[(-0.898) 2 + (-4.898) 2] = 4.98
Θ = tan-1 (4.898/0.898) = 79.6 o = 1.389
r = 4.98 ( cos 79.6 + sin 79.6)
r ^1/3 =[ 4.98ei 1.389]1/3 = 4.98 1/3 x [ cos((360n + 79.6)/3)) + sin((360n + 79.6) /3)]
n =0, r 1/3 = 1.71 x (0.895 + 0.447i) = 1.53 + 0.764 i
n = 1, r 1/3 = 1.71 x (- 0.834 + 0.551 i) = 1.71 (cos 113.2 + sin 113.2) = -1.426 + 0.943 i
n = 2, r 1/3 = 1.71 x (-0.060 - 0.998i) = -1.03 – 1.707 i

 

Answer: 1.71 (cos 113.2 + sin 113.2) = -1.426 + 0.943 i

 

Check over the whole solution to ensure there are not calculation errors. I was in a hurry when I did this.