Question

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

Answer 1

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

`|z|=√((-\sqrt3)^2+1^2)=2`

`z` lies in the second quadrant, so

`\arg z=\pi+\tan^(-1)\left(-\frac1{\sqrt3}\right)=\frac{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)`