4th root question/guidance Find all 4th roots of $-8 + 8i\sqrt 3$
$a=-8$
$b=8\sqrt 3$
$r= \sqrt{a^2+b^2}= \sqrt {(-8^2)+(8\sqrt{3})^2)}=\sqrt{64+192}=\sqrt {256} =16$
$\frac ar= cos\theta=\frac{-1}{2}$     $\space     $ $\frac br= sin\theta$=$\frac {\sqrt3}{2}$  
This gives me a different $\theta$ one being 120 degrees and the other 60 degrees
I can get the roots I just don't see were I am making the mistake with $\theta$, after that point I can finish it off myself. 

Polar form:
$z_1= 2(\cos 30 + i \sin 30)$
$z_2= 2(\cos 120 + i \sin 120)$
$z_3= 2(\cos 210 + i \sin 210)$
$z_4= 2(\cos 300 + i \sin 300)$
Rectangular Form:
$z_1= 2(\frac {\sqrt{3}}{2} + i \frac 12) = \sqrt {3} + i$ 
$z_2= 2(-\frac 12 + i\frac {\sqrt{3}}{2} ) =-1 +i \sqrt {3}$
$z_3= 2(-\frac {\sqrt{3}}{2} - i \frac 12)=-\sqrt {3}  -i$
$z_4= 2(\frac 12 - i\frac {\sqrt{3}}{2}) = 1 -i\sqrt {3}$
