Let [tex]z=-\sqrt3+i[/tex]. Then
[tex]|z|=\sqrt{(-\sqrt3)^2+1^2}=2[/tex]
[tex]z[/tex] lies in the second quadrant, so
[tex]\arg z=\pi+\tan^{-1}\left(-\dfrac1{\sqrt3}\right)=\dfrac{5\pi}6[/tex]
So we have
[tex]z=2e^{i5\pi/6}[/tex]
and the fourth roots of [tex]z[/tex] are
[tex]2^{1/4}e^{i(5\pi/6+k\pi)/4}[/tex]
where [tex]k\in\{0,1,2,3\}[/tex]. In particular, they are
[tex]2^{1/4}e^{i(5\pi/6)/4}=2^{1/4}e^{i5\pi/24}[/tex]
[tex]2^{1/4}e^{i(5\pi/6+2\pi)/4}=2^{1/4}e^{i17\pi/24}[/tex]
[tex]2^{1/4}e^{i(5\pi/6+4\pi)/4}=2^{1/4}e^{i29\pi/24}[/tex]
[tex]2^{1/4}e^{i(5\pi/6+6\pi)/4}=2^{1/4}e^{i41\pi/24}[/tex]
