A regular hexagon is drawn with two of its vertices forming a shorter diagonal at z=−2  and z=1−i3 . The other four vertices are:

AC=23=2rsin600=2 |OA|=2⇒O  is origin.∴ z4=2z2−0−2−0=e−π/3=12+i32⇒    z2=−1−i3∴    z5=1+i3Also−2−0z6−0=eiπ/3=12+i32⇒   z6=−41+i3=−4(1−i3)3=−1+i3∴    Remaining vertices are2,  −i+i3,  1+i3,  −1−i3