+499 Here we make use of de moivre's theorem, which states that given a complex number z:
\(z=re^{i\theta} = r(\cos{\theta} + i\sin{\theta})\)
We can then raise z to the nth power to get a general form for this:
\(z^n = r^ne^{i*n*\theta} = r(\cos{n\theta} + i\sin{n\theta})\)
We can then set z as our desired complex number. We then get:
\(-3+3i\sqrt{3} = 6(\cos{2\pi\over3} + i\sin{2\pi\over3})\)
More generally, we can write:
\(-3+3i\sqrt{3} = 6(\cos(2n\pi +{2\pi\over3}) + i\sin(2n\pi + {2\pi\over3}))\)
\((-3+3i\sqrt{3})^{1/4} = 6(\cos(2n\pi/4 +{2\pi\over12}) + i\sin(2n\pi/4 + {2\pi\over12}))\)
Simplifying, we get:
\((-3+3i\sqrt{3})^{1/4} = 6(\cos(n\pi/2 +{\pi\over6}) + i\sin(n\pi/2 + {\pi\over6}))\)
Now we get our 4 roots from testing n from 0-3(it iterates after a cycle of 4). I'm sure you're capable enough to plug and chug in the numbers! :)
Edit: here's a good reference; it uses a similar method and explanation
https://socratic.org/questions/how-do-you-find-the-fourth-roots-of-i
