Let I : Arg((z-8i)/(z+6)) = +- pi/2
II: Re((z-8i)/(z+6)) = 0
Show that locus of z in I or II lies on x^2 + y^2 + 6x - 8y = 0. Hence show that locus of z can also be represented by (z-8i)/(z+6) + (z'+8i)/(z'+6) = 0. Further if locus of z is expressed as |z + 3 - 4i| = R, then find R.