Question

Find the complex fourth roots \[-\sqrt{3}+\iota \] in polar form.

Answer 1

Let $z=-\sqrt3+i$. Then

$|z|=\sqrt{(-\sqrt3)^2+1^2}=2$

$z$ lies in the second quadrant, so

$\arg z=\pi+\tan^{-1}\left(-\dfrac1{\sqrt3}\right)=\dfrac{5\pi}6$

So we have

$z=2e^{i5\pi/6}$

and the fourth roots of $z$ are

$2^{1/4}e^{i(5\pi/6+k\pi)/4}$

where $k\in\{0,1,2,3\}$. In particular, they are

$2^{1/4}e^{i(5\pi/6)/4}=2^{1/4}e^{i5\pi/24}$

$2^{1/4}e^{i(5\pi/6+2\pi)/4}=2^{1/4}e^{i17\pi/24}$

$2^{1/4}e^{i(5\pi/6+4\pi)/4}=2^{1/4}e^{i29\pi/24}$

$2^{1/4}e^{i(5\pi/6+6\pi)/4}=2^{1/4}e^{i41\pi/24}$