<s>[INST] 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.
 [/INST] 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
$\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i} = \frac{1}{8} \pm \frac{3\sqrt 7}{8}i$
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