If #(z_1)^2+(z_2)^2+(z_3)^2-z_1z_3-z_3z_2-z_1z_2=0#, prove that,#|z_2-z_3|=|z_3-z_1|=|z_1-z_2|#, where #z_1,z_2,z_3# are complex numbers?

Question

2747 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-03T13:11:25

See below.

Making

#u = z_1#
# v = z_2#
#w = z_3#

calling

#m = u-v#
#n = v-w#
#p = u-w#

we have

#m^2+n^2+p^2 = 2(u^2+v^2+w^2-u*v-u*w-v*w)#

but if

#u^2+v^2+w^2-u*v-u*w-v*w = 0# ten

#m^2+n^2+p^2 =0 rArr abs(u-v)=abs(v-w)=abs(u-w)#