Complex numbers - finding a square root of something Let $z_1 , z_2 $ be two complex numbers that satisfy:
$\dfrac{z_2 } {\bar{z_1}}= \frac{3}{8} \big(\cos(75^{\circ})+i\sin(75^{\circ})\big) $ , 
$z_1 z_2 ^2 = \frac{1}{3} \big(\cos(120^{\circ}) + i\sin(120^{\circ}) \big) $ . 
How can I determine with of  the following can be a possible value for $ \sqrt{z_1} $ ?  
(a) $ \frac{2}{\sqrt{3}} \mbox{cis}(135^{\circ}) $ 
(b) $ \frac{2}{3} \mbox{cis}(155^{\circ})$
(c)  $ \frac{2}{\sqrt{3}} \mbox{cis}(195^{\circ}) $ 
(d) $ \frac{2}{\sqrt{3}} \mbox{cis}(175^{\circ}) $ 
(e) $ \frac{2}{3} \mbox{cis}(215^{\circ})$
thanks !!! 
 A: Write the $z_i$ in polar representation as $z_1 = r_1 \mathrm{cis}(\theta_1)$ and $z_2 = r_2 \mathrm{cis}(\theta_2)$. Then, the two equations you have are:
$$ \frac{z_2}{\bar{z_1}} = (r_2 \mathrm{cis}(\theta_2))(\frac{1}{r_1} \mathrm{cis}(\theta_1)) = \frac{r_2}{r_1}\mathrm{cis}(\theta_1 + \theta_2) = \frac{3}{8}\mathrm{cis}(75^{\circ}), $$
$$ z_1 z_2^2 = r_1 r_2^2 \mathrm{cis}(\theta_1 + 2\theta_2) = \frac{1}{3} \mathrm{cis}(120^{\circ}).$$
What does the equations imply on $r_i$ and $\theta_i$?
A: If you have: z1 = a*[$\cos (x_1)$ + i*$\sin (x_1)$] and z2 = b*[$\cos (x_2)$ + i*$\sin (x_2)$], then you have: 
$ z_2/z_1$ = a/b * [$\cos (x_2-x_1)$ + i*$\sin (x_2-x_1)$] , and from this equation and your first one you`ll have: 
$a/b$ = 3/8 and x2-x1= 75.
$z_2^2$ = $b^2$ [$\cos (2x_2)$ + i*$\sin (2x_2)$] (Moivre) and you have:
$z_1*z_2^2$= $a*b^2$ [$\cos (x_1+2x_2^2)$ + i$\sin (x1+2x2^2)$];  so from the second equation result that :
$a*b^2$ = 1/3; 
x1+ 2*x2= 120;
With a/b = 3/8 and x2-x1= 75.
We have 3*x2=195 so x2= 65, then x1 = -10 , and a= 4/3 
So x1 = 4/3 [$\cos (350)$ + i*$\sin (350)$]
For the square root of x you have:

So: we hace $x_3$ = c * [$\cos (x)$ + i*$\sin (x)$] as the root of x1
From the formula we have c= $2/\sqrt3$. and x= 350/2 =175. 
$x_3$ = $2/\sqrt3$ * [$\cos (175)$ + i*$\sin (175)$]
Final result D
***Sorry for the writing, I am new , and I am now learning haow to write mathematics symbol in stackexchange!
