If a complex number z is such that z-2 i z-2 purely imaginary number and the locus of z is a closed curve, then the area of the region bounded by that closed curve and lying in the first quadrant is

Step-by-step solution

Given, z-2 i z-2 is purely imaginary. aligned & z-2 i z-2 + ( z-2 i z-2 )=2 Re ( z-2 i z-2 ) & z-2 i z-2 + z +2 i z -2 =0 & |z|^2-2 z-2 i z +4 i+|z|^2+2 i z-2 z -4 i= & 2|z|^2-2(z+ z )+2 i(z- z )=0 & x^2+y^2-2 x-2 y=0 ( z=x+i y) & (x-1)^2+(y-1)^2=( 2 )^2 & Required area = r^2 2 = ( 2 )^2 4 = 2 aligned