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Find all $x \in \mathbb{C}$ satisfying $(x - \sqrt{3} + 2i)^3 - 8i = 0$

I was able to find one value, $x = \sqrt{3} - 4i$.

I can also see that $x = -i$ also works although I am not able to devise a formal way for ending up with this value.

I am also unsure of how to find other values of $x$.

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    $\begingroup$ Hint: let $w=x-\sqrt{3}+2i\,$, then solve $w^3 = 8i$ for $w$. $\endgroup$ – dxiv Mar 29 '17 at 23:48
  • $\begingroup$ Once you find one root, you can find the other two by quadratic formula. $\endgroup$ – Yunus Syed Mar 29 '17 at 23:48
  • $\begingroup$ You can even apply a sum/difference of cubes formula here. But @dxiv gives probably the most efficient route. $\endgroup$ – pjs36 Mar 29 '17 at 23:51
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Hint: Divide your polynomial by $(x-(\sqrt{3} - 4i))$. You will get a quadratic polynomial. You found that $x = -i$ is also a root (great work!) so you should be to divide again by $(x+i)$.

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Rewrite the equation as $$(x-\sqrt{3}+2i)^3=8i=(-2i)^3,$$ to find that $$\left(\frac{x-\sqrt{3}+2i}{-2i}\right)^3=1.$$ So you're looking for roots of $w^3-1=(w-1)(w^2+w+1)$, where $w=\frac{x-\sqrt{3}+2i}{-2i}$.

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.Rewrite the equation as : $(x-\sqrt3 + 2i)^3 = 8i$.

Express $8i$ in polar form, you can write it as $8 e^{\frac{i\pi}{2}}$.

Therefore, this question is basically asking you : what are the complex cube roots of $8i$? Thankfully, these can be answered via a use of De Moivre's theorem, which will tell you that the roots are precisely those of the form $\sqrt[3]{8} e^{\frac {i\pi}{6} + \frac{2in \pi}{3}}$ where $n=0,1,2$.

Now, this can be simplified: $$ \sqrt[3]{8} e^{\frac \pi{6} + \frac{2n \pi}{3}} = 2 \operatorname{cis} \frac \pi {6}, 2 \operatorname{cis} \frac {5\pi} {6}, 2 \operatorname{cis} \frac {9\pi} {6} $$

Expand using the fact that $\operatorname{cis} \theta = \cos \theta + i \sin \theta$, and don't forget to subtract $2i - \sqrt 3$ from each of the cube roots. This gives an answer that involves only sine and cosine evaluation. Alternately, since you figured out a root, you could also have done polynomial division, but for that you would have to expand the cube root, which is laborious.

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