If `|z_1-z_0|=|z_2-z_0|=a` and `amp((z_2-z_0)/(z_0-z_1))=pi/2` , then find `z_0`

If |z_1| = |z_2| =1 and amp z_1 +amp z_2 =0 then

if |z_1| = |z_2| ne 0 and amp (z_1)/(z_2) = pi then

If |z_(1)|=|z_(2)|andamp(z_(1))+amp(z_(2))=0, then

If |z_1+z_2| = |z_1-z_2| , prove that amp z_1 - amp z_2 = pi/2 .

Statement I: If |z_1+z_2|=|z_1|+|z_2|, then Im(z_1/z_2)=0 (z_1,z_2 !=0) Statement II: If |z_1+z_2|=|z_1|+|z_2| then origin, z_1, z_2 are collinear with 'z_1' and z_2 lies on the same side of the origin (z_1,z_2 !=0)

If z_1,z_2,z_3,z_4 be the vertices of a parallelogram taken in anticlockwise direction and |z_1-z_2|=|z_1-z_4|, then sum_(r=1)^4(-1)^r z_r=0 (b) z_1+z_2-z_3-z_4=0 a r g(z_4-z_2)/(z_3-z_1)=pi/2 (d) None of these

If |z-i|=1 and amp(z)=(pi)/(2)(z!=0), then z is

If |z_(1)|=|z_(2)|=|z_(3)| and z_(1)+z_(2)+z_(3)=0 , then z_(1),z_(2),z_(3) are vertices of

If |z_1 |=|z_2|=|z_3| = 1 and z_1 +z_2+z_3 =0 then the area of the triangle whose vertices are z_1 ,z_2 ,z_3 is