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.
