If the amplitude of `((z-2)/(z-6i))=(pi)/(2)` find its locus.

Explanation

Let `z=(x+iy)`
Then `(z-2)/(z-6i)=(x+iy-2)/(x+iy-6i)=((x-2)+iy)/(x+i(y-6))`
`([(x-2)+iy][x-i(y-6)])/([x+i(y-6)][x-i(y-6)])`
`([(x-2)+y(y-6)])/(x^(2)+(y-6)^(2))+i(xy-(x-2)(y-2))/(x^(2)+(y-6)^(2))`
`=a+ib`, say then
`a=(x(x-2)+y(y-6))/(x^(2)(y-6)^(2))`
`b=(xy-(x-2)(y-2))/(x^(2)+(y-6)^(2))`
By hypothesis, amplitude of `a+ib=(pi)/(2)`
So `(pi)/(2)= "tan"^(-1)(b)/(a)`
Hence `a=0 and b gt 0`
`:.x(x-2)+y(y-6)=0`
` or x^(2)+y^(2)-2x-6y=0`