If z=x+i y, where x and y are real numbers and i= -1 , then the points (x, y) for which z-1 z-i is real, lie on

Step-by-step solution

Given, z=x+i y Now, z-1 z-i = (x+y)-1 (x+i y)-i = (x-1)+i y x+i(y-1) x-i(y-1) x-i(y-1) = x(x-1)+i x y-i(x-1)(y-1)+y(y-1) x²+(y-1)² = (x²+y²-x-y )+i(x y-x y+y+x-1) (x²+y²+1^ -2 y ) = ( x²+y²-x-y x²+y²-2 y+1 )+i ( x+y-1 x²+y²-2 y+1 ) (0) Given, z-1 z-i is real. So, its imaginary part should be zero. i.e.. x+y-1 x²+y²-2 y+1 =0 x+y=1 ( x²+y²-2 y+1 0 ) Which represent a straight line.