# Roots of Complex Polynomials

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 $x$.

• Hint: let $w=x-\sqrt{3}+2i\,$, then solve $w^3 = 8i$ for $w$. – dxiv Mar 29 '17 at 23:48
• Once you find one root, you can find the other two by quadratic formula. – Yunus Syed Mar 29 '17 at 23:48
• You can even apply a sum/difference of cubes formula here. But @dxiv gives probably the most efficient route. – pjs36 Mar 29 '17 at 23:51

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 $(x+i)$.

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 $w=\frac{x-\sqrt{3}+2i}{-2i}$.

.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.