If `z=z_0+A( bar z -( bar z _0)), w h e r eA` is a constant, then prove that locus of `z` is a straight line.

If z=z_(0)+A(bar(z)-(bar(z)_(0))), whereA is a constant,then prove that locus of z is a straight line.

The equation bar(z)=bar(z_(0))+A(z-z_(0)) where A is a constant,and if m is the slope of the straight line then A is

If |z+bar(z)|= |z-bar(z)| , then the locus of z is

If arg (bar(z)_(1))= "arg" (z_(2)), (z ne 0) then

If |z+ bar(z)|+ |z-bar(z)| =2 , then z lies on

If z ne0 , find Arge z+Arg bar z