Trigonometrical solution of complex equation Need present in trigonometrical form the solution of the complex equation: $x^6 = 1 + \sqrt3 + (1-\sqrt3)i$ To take out the coefficients of the real & imaginary parts real part: $1 + \sqrt3$, imaginary part: $(1-\sqrt3)$ Now, for converting into polar form, $r = \sqrt8$; but the other part of finding the appropriate angle representing the coefficients is not possible. Any hint will be most welcome.

So, you know that $\cos\theta=\frac{1+\sqrt3}{2\sqrt2}$ and that $\sin\theta=\frac{1-\sqrt3}{2\sqrt2}$. Therefore, $2\sin(\theta)\cos(\theta)=-\frac12$, which means that $\sin(2\theta)=\sin\left(-\frac\pi6\right)$. This suggests (and it is easy to prove) that $\theta=-\frac\pi{12}$. So$$x^6=2\sqrt2\left(\cos\left(-\frac\pi{12}\right)+\sin\left(-\frac\pi{12}\right)i\right).$$Can you take it from here?