Reducing the complex expression $\sqrt[6]{\frac{1-i}{\sqrt 3 + i}}$

Question

By multiplying the numerator and the denominator by the conjugate of the denominator, I managed to reduce the expression down to the following:

$$\sqrt[6]{\frac{1-i}{\sqrt 3 + i}} = \sqrt[6]{\frac{(1-i)(\sqrt 3 - i)}{(\sqrt 3 + i)(\sqrt 3 - i)}} = \sqrt[6]{\frac{\sqrt{3}-1+i\left(-\sqrt{3}-1\right)}{4}}$$

Now I'm stuck. I don't know how to convert this expression to a trigonometric form without using a calculator, so I can't apply De Moivre's formula to get all the roots. Is there a way to continue the simplification?

Answer 1

Hint: Convert the numerator and denominator of the radicand to polar form first.

Answer 2

$$\sqrt[6]{\frac{1-i}{\sqrt 3 + i}}=\sqrt[6]{\frac{\sqrt2(\cos315^{\circ}+i\sin315^{\circ})}{2(\cos30^{\circ}+i\sin30^{\circ})}}=\frac{1}{\sqrt[12]2}\sqrt[6]{\cos285^{\circ}+i\sin285^{\circ}}$$ Can you end it now?