# Solving $z^4=(\frac{1-\sqrt{3}i}{1+\sqrt{3}i})^3$ using De moivre's

$$z^4=\left(\frac{1-\sqrt{3}i}{1+\sqrt{3}i}\right)^3$$

I was trying to solve it using De Moivre's and I am stuck

$z^4 = (\frac{e^{-i\pi/3}}{e^{i\pi/3}})^3 = (e^{-2i\pi/3})^3 = e^{-2i\pi}$.

Write $e^{-2i\pi} = 1 = e^{2i\pi} = e^{4i\pi}$ and take fourth root of each term to obtain the roots as $e^{-i\pi/2},1,e^{i\pi/2},e^{i\pi}$ which are $-i,1,i,-1$ respectively.

$$z^4=\left(\frac{1-\sqrt{3i}}{1+\sqrt{3i}}\right)^3\Longleftrightarrow$$ $$z^4=\left|\left(\frac{1-\sqrt{3i}}{1+\sqrt{3i}}\right)^3\right|e^{\arg\left(\left(\frac{1-\sqrt{3i}}{1+\sqrt{3i}}\right)^3\right)i}\Longleftrightarrow$$ $$z^4=5\sqrt{\frac{5}{4499+1836\sqrt{6}}}e^{-\tan^{-1}\left(\frac{3\sqrt{\frac{3}{2}}}{7}\right)i}\Longleftrightarrow$$ $$z=\left(5\sqrt{\frac{5}{4499+1836\sqrt{6}}}e^{\left(-\tan^{-1}\left(\frac{3\sqrt{\frac{3}{2}}}{7}\right)+2\pi k\right)i}\right)^{\frac{1}{4}}$$

With $k\in\mathbb{Z}$ and $k:0-4$

Make it more likeable with your own math skills!

Hint

After manipulations, you can arrive to $$\frac{1-\sqrt{3i}}{1+\sqrt{3i}}=\frac{1}{5} \left(\sqrt{6}-4\right)+i \frac{1}{5} \left(3-2 \sqrt{6}\right)$$ Rising to cube and simplifying again $$\left(\frac{1-\sqrt{3i}}{1+\sqrt{3i}}\right)^3=\frac{28-6 i \sqrt{6}}{\left(4+\sqrt{6}\right)^3}$$

I am sure that you can take from here.