Evaluating $\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$ Let $A(z_1),B(z_2),C(z_3)$ are complex numbers satisfying $|z-\sqrt{3}i|=1$ and $3z_1+\sqrt{3}i=2z_2+2z_3$. The question asks to find the value of $$\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$$ I tried to shift the origin to $\sqrt{3}i$ and transformed the original condition in this system but it does not help me. Any ideas how to proceed? Thanks.
 A: We can rewrite the equation as $3(z_1 - \sqrt {3} i) = 2(z_2-\sqrt {3}i)+2(z_3-\sqrt {3}i)$
i.e. $3\vec{OZ_1} =  2\vec{OZ_2}+2 \vec{OZ_3}$
So if $\theta = \pm \arg \frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$ is the angle between $\vec{OZ_2}$ and $\vec{OZ_3}$, we have from the above equation that
$9 = 2^2+2^2+8 \cos \theta \Rightarrow \cos \theta = \frac{1}{8} \Rightarrow \sin \theta = \pm \frac{3\sqrt 7}{8}$
Since $\left| \frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}\right|=1$, we have that
$\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i} = \frac{1}{8} \pm \frac{3\sqrt 7}{8}i$
A: Put $t_j = z_j - \sqrt{3}i$. Then $|t_j|=1$ and $3t_1 = 2t_2 + 2t_3$. Hence $\frac{3}{4}t_1 = \frac{1}{2}(t_2 + t_3)$. The points $t_j$ lie on the unit circle, with center $O$ at the origin and form a triangle $ABC$, with $A = t_1$ etc. The midpoint $D$ of the chord  $BC$ is $\frac{3}{4}t_1$ and hence lies on $OA$ at a distance $\frac{3}{4}$. This means that in the right angled triangle $ODB$, $OB = 1, OD = \frac{3}{4}$. Thus if the angle $BOC$ is $2\theta$, we have $\cos\theta = \frac{3}{4}$. Now $\frac{t_2}{t_3} = \cos 2\theta + i \sin 2\theta$ and $\cos 2\theta = 2\cos^2 \theta -1 = \frac{1}{8}$ and $\sin 2\theta = \pm\sqrt{1-\cos^2 2\theta} = \pm\frac{\sqrt{63}}{8}$. Thus the required ratio is 
$$\frac{1}{8} \pm 3\frac{\sqrt{7}}{8} i$$ 
