Although the answers are requested in rectangular (a + bi) form, exponentiation (including roots, aka fractional exponentiation) of complex numbers is more easily done in polar form. Thus, we can find the solutions in polar form, and convert them into a+bi form after:
A) 36 = 729, so all of the 6th roots of -729 will have a norm (modulus) of 3. One such evident solution is 3i, since i6 = -1. When we make an Argand diagram for 3i (graph it in the complex plane), it lies 3 units due "West" of the origin. Thus its polar form is 3cis(π/2), shorthand for 3[cos(π /2)+ isin(π/2)]. The remaining 5 roots lie evenly spaced in the complex plane, so 60° or π/3 apart: 3cis(π/6) , 3cis(5π/6) , 3cis(7π/6) , 3cis(3π/2) , and 3cis11π/6). In rectangular form: 3√3/2 + 3/2i , 3i , -3√3/2 + 3/2i , -3√3/2 - 3/2i , - 3i , and 3√3/2 - 3/2i.
B) Similar to the above, since x3 = - 8i, and since -8i = 8cis3π/2 , one cube root is 2cisπ/2 (we cube root the modulus and divide the angle by 3), or 2i. The remaining cube roots lie 120º apart in the complex plane (forming an equilateral triangle centered on the origin, 0 +0i). Thus, the two remaining solutions are 2cis7π/6 and 2cis11π/6. In a + bi form, the 3 solutions are 2i , - 1 - √3i , and 1 - √3i.
We can confirm the last solution easily enough by hand: (1 - √3i)3 = 13 + 3·12·√3i + 3·1·(√3i)2 + (√3i)3 = 1 + 3√3i - 9 - 3√3i = - 8 .
