If `|z^2-1|=|z|^2+1` show that the locus of z is as straight line.

Statement-1: If z is a complex number satisfying (z-1)^(n) , n in N , then the locus of z is a straight line parallel to imaginary axis. Statement-2: The locus of a point equidistant from two given points is the perpendicular bisector of the line segment joining them.

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 be any complex number such that |3z-2|+|3z+2|=4 , then show that locus of z is a line-segment.

If z^2+z|z|+|z^2|=0, then the locus z is a. a circle b. a straight line c. a pair of straight line d. none of these

Statement-1 : The locus of z , if arg((z-1)/(z+1)) = pi/2 is a circle. and Statement -2 : |(z-2)/(z+2)| = pi/2 , then the locus of z is a circle.

If the real part of (z +2)/(z -1) is 4, then show that the locus of he point representing z in the complex plane is a circle.

If the imaginary part of (2z+1)/(i z+1) is -2 , then show that the locus of the point respresenting z in the argand plane is a straight line.

If |z|=1 and let omega=((1-z)^2)/(1-z^2) , then prove that the locus of omega is equivalent to |z-2|=|z+2|

If Imz((z-1)/(2z+1))=-4 , then locus of z is

If |z| = 2 and the locus of 5z-1 is the circle having radius 'a' and z_1^2 + z_2^2 - 2 z_1 z_2 cos theta = 0 then |z_1| : |z_2| =