The fourth roots of $-8+i\cdot8\sqrt3$.

$$-8+i\cdot8\sqrt3$$ converted to polar form is $$16 \exp(\pi/3)$$.

According to the theory,

fourth root of $$16 = 2$$, so all four roots should be calculated from:

$$2\cdot\exp(i\cdot(\pi/12 + 2\pi k/4))$$ where $$k=0,1,2,3$$.

I'm struggling to get the right answers from that. i.e. one of the answers should be: $$\sqrt3 - i.$$

• Your polar form is not correct. Note that the real part is negative. – xbh Mar 3 '19 at 1:27

As alluded to in the comments, $$-8+8\sqrt3i = 16 e^{2\pi i/3}.$$ Therefore, its fourth roots are $$2e^{i(\pi/6+2\pi k/4)} =2e^{i\pi(1/6+k/2)}=2e^{i\pi(1+3k)/6}$$ with $$k \in \{0,1,2,3\},$$ i.e., $$2e^{i\pi/6},$$ $$2e^{i\pi 2/3},$$ $$2e^{i\pi7/6},$$ and $$2e^{i\pi5/3 },$$ or in non-polar notation
$$2({\sqrt3 \over 2}+{1 \over 2}i)= \color{red}{\sqrt3+i}, \; 2({-1 \over 2} + {{\sqrt3} \over 2}i)=\color{red}{-1+\sqrt3i},\; 2({{-\sqrt3}\over2}+{{-1}\over2}i)=\color{red}{-\sqrt3-i},$$ and
$$2(\frac 1 2 + {{ -\sqrt 3}\over 2} i)=\color{red}{1-\sqrt3i}.$$
Note that $$(\sqrt3-i)^4=-8-i\cdot8\sqrt3.$$
• @NapoleonCornejo: when you wrote that one of the answers should be $\sqrt3 -i$, did you mean $-\sqrt3-i$ or $\sqrt3+i$? – J. W. Tanner Mar 3 '19 at 14:48