If the points lie on a circle with radius 1 then they can be represented as
e^(i phi_1)+e^(i phi_2)+e^(i phi_3)+e^(i phi_4)=0
or equivalently
sin phi_1+sin phi_2+sin phi_3+sin phi_4 = 0
and
cos phi_1+cos phi_2+cos phi_3+cos phi_4 = 0
(Here we used de Moivre's identity e^(i phi) = cos phi + i sin phi )
grouping and squaring both sides
(sin phi_1+sin phi_2)^2=(-(sin phi_3+sin phi_4))^2
(cos phi_1+cos phi_2)^2=(-(cos phi_3+cos phi_4))^2
and adding side by side we get at
2+2(sin phi_1 sin phi_2+cos phi_1 cos phi_2) = 2 +2(sin phi_3 sin phi_4+ cos phi_3 cos phi_4)
or
cos(phi_2-phi_1) = cos(phi_4-phi_3)
grouping now
(sin phi_2+sin phi_3)^2=(-(sin phi_1+sin phi_4))^2
(cos phi_2+cos phi_3)^2=(-(cos phi_1+cos phi_4))^2
we get at
cos(phi_3-phi_2) = cos(phi_4-phi_1)
So we can conclude that the quadrilateral is a regular quadrilateral.
