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$-8+i\cdot8\sqrt3$ 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: $\sqrt3 - i.$

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    $\begingroup$ Your polar form is not correct. Note that the real part is negative. $\endgroup$ – xbh Mar 3 '19 at 1:27
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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 $(\sqrt3-i)^4=-8-i\cdot8\sqrt3.$

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  • $\begingroup$ Yes, I realized my mistake was actually missing the proper angle from the arctan. The proper angle is as you state (2pi/3). It's clear now, Thanks! $\endgroup$ – Napoleon Cornejo Mar 3 '19 at 10:10
  • $\begingroup$ @NapoleonCornejo: when you wrote that one of the answers should be $\sqrt3 -i$, did you mean $-\sqrt3-i$ or $\sqrt3+i$? $\endgroup$ – J. W. Tanner Mar 3 '19 at 14:48

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