If $A(z_1) , B(z_2) , C(z_3)$ are complex numbers satisfying $|\mathrm{z}-\sqrt{3} \mathrm{i}|=1$ and .... Find $\left|z_{1}-z_{2}\right|$ If $A(z_1) , B(z_2) , C(z_3)$ are complex numbers satisfying $|\mathrm{z}-\sqrt{3} \mathrm{i}|=1$ and $3 \mathrm{z}_{1}+\sqrt{3} \mathrm{i}=2 \mathrm{z}_{2}+2 \mathrm{z}_{3}$. Find $\left|z_{1}-z_{2}\right|$
My Approach: I can more or less just figure out that these complex numbers correspond to point on the circle $x^2 + (y-\sqrt(3))^2 = 1$. I have absolutely no idea how to proceed after that. I cant think of a geometrical solution nor otherwise. Please help me in understanding how I should proceed from here?
Any hints/explanations are helpful and appreciated. Thanks!
EDIT: The question does say $|z1-z2|$ with an answer of $\frac{1}{\sqrt{2}}$ but it could be a misprint aswell. Please check if $|z2-z3|$ satisfies your solution it might be what they were looking for.
 A: To simplify the calculations based on the circle condition, let $u = z - i\sqrt{3}$, then $\require{cancel}|u|=1$ and $3\left(u_1+\cancel{i\sqrt{3}}\right) + \cancel{i \sqrt{3}} = 2\left(u_2+\cancel{i\sqrt{3}}\right)+2\left(u_3+\cancel{i\sqrt{3}}\right) \iff 3 u_1 = 2(u_2+u_3)\,$.
( As a side note and to explain my comment, this implies $|u_2+u_3|=\frac{3}{2}$, so it immediately follows by the parallelogram law that $|u_2-u_3|^2 = 2\left(|u_2|^2+|u_3|^2\right) - |u_2+u_3|^2 = \frac{7}{4}\,$. )
Back to the problem as posed, the geometric insight here is that $u_2+u_3$ is collinear with $u_1$ which, given the unit circle constraint, means $u_2$ and $u_3$ are symmetric with respect to the $u_1$ direction, so that $u_1, u_2, u_3$ and the origin are the vertices of a kite, and therefore $|u_1-u_2|=|u_1-u_3|$.
To formalize it algebraically, let $v_2=\frac{u_2}{u_1}$ and $v_3=\frac{u_3}{u_1}$, then $|v_2|=|v_3|=1$ and $v_2+v_3=\frac{3}{2}$. Writing it as $v_3 = \frac{3}{2} - v_2$ and multiplying by the conjugate gives: $$1 = |v_3|^2 = \left(\frac{3}{2}-v_2\right)\left(\frac{3}{2}-\bar v_2\right) = \frac{9}{4}+|v_2|^2-\bcancel{2} \cdot \frac{3}{\bcancel{2}}\cdot\Re{(v_2)} \;\;\iff\;\; \Re{(v_2)}=\frac{3}{4}$$
It follows that $v_2 = \dfrac{3 \pm i \sqrt{7}}{4}$, then: $$|z_1-z_2|=|u_1-u_2|=\left|\frac{u_1-u_2}{u_1}\right|=\left|1-v_2\right|=\left|\frac{1\pm i\sqrt{7}}{4}\right|=\frac{1}{\sqrt{2}}$$
