Complex number geometry

The mid point of $z_{2}~and ~z_{3}$ divides the line joining $z_{1}$ and $i\sqrt{3}$ in the ratio $1:3$ internally.

$\frac{3z_{1}+i\sqrt{3}}{4}=\frac{z_{2}+z_{3}}{2}$

$A,B,C$ lie on the circle with centre $(0,\sqrt{3})$ and radius of $1$ unit in the argand plane .

$|DO|=|\frac{3AO}{4}|~sand~|AD|=|\frac{AO}{4}|$

$|AO|=|BO|=|CO|=1$ . using pythagoras theorem in right triangle $BDO$

$|BD|=\frac{\sqrt{7}}{4}$

using pythagoras theorem in right triangle $BAD$

$\large{|AB|=|z_{1}-z_{2}|=\frac{1}{\sqrt{2}}}$

$|z-i\sqrt{3}|=1 \implies z-i\sqrt{3}=e^{i\theta}$ Now, $3z_{1}+i\sqrt{3}=2z_{2}+z_{3} \implies 3i\sqrt{3}+3e^{i\theta_{1}}+i\sqrt{3}=2[2i\sqrt{3}+e^{i\theta_{2}}+e^{i\theta_{3}}]$ Taking the real and imaginary part separately we get $3\sin{\theta_{1}}-2\sin{\theta_{2}}=2\sin{\theta_{3}}$ and $3\cos{\theta_{1}}-2\cos{\theta_{2}}=2\cos{\theta_{3}}$ Now take square of these two equations and add them . Then we get, $\cos{(\theta_{1}-\theta_{2})}=\frac{3}{4}$ Now $|z_{1}-z_{2}|=|e^{i\theta_{1}}-e^{i\theta_{2}}|$ $=\sqrt{2-2\cos{[\theta_{1}-\theta_{2}]}}$ $=\sqrt{2-2\times\frac{3}{4}}$ $=\frac{1}{\sqrt{2}}=0.707$