A regular hexagon is drawn with two of its vertices forming a shorter diagonal at $z=-2$  and $z=1-i\sqrt{3}$ . The other four vertices are:

$AC=2\sqrt{3}=2r\sin {60}^0=2$ 

$\left|OA\right|=2\Rightarrow O$  is origin.

$\therefore$ $z_4=2$

$\frac{z_2-0}{-2-0}=e^{-\pi /3}=\frac{1}{2}+i\frac{\sqrt{3}}{2}$

$\Rightarrow$    $z_2=-1-i\sqrt{3}$

$\therefore$    $z_5=1+i\sqrt{3}$

Also$\frac{-2-0}{z_6-0}=e^{i\pi /3}=\frac{1}{2}+i\frac{\sqrt{3}}{2}$

$\Rightarrow$   $z_6=\frac{-4}{1+i\sqrt{3}}=\frac{-4\left(1-i\sqrt{3}\right)}{3}=-1+i\sqrt{3}$

$\therefore$    Remaining vertices are

$2,-i+i\sqrt{3},1+i\sqrt{3},-1-i\sqrt{3}$