Let `(2-i)z=(2+i)barz` and `(2+i)z +(-2+i)barz-i=0` be normal to the circle and `iz+barz+1+i=0` is tangent to the same circle having radius r . Then vaue of `128r^2`

Let the lines (2-i)z=(2+i) barz and (2+i)z+(i-2)barz-4i=0 , (here i^(2)= -1 ) be normal to a circle C. If the line iz+barz+1+i=0 is tangent to this circle C, then its radius is :

The center of the circle . zbarz+(1+i)z +(1+i)barz-7=0 are respectively.

The radius of the circle |(z-i)/(z+i)|=3 , is

If (1+i)z=(1-i)barz , then z is equal to

If (1+i)z = (1-i)barz, "then show that "z = -ibarz.

Find the complex number z if z^2+barz =0

The equation zbarz+(4-3i)z+(4+3i)barz+5=0 represents a circle of radius