Question

Prove that locus of $z$ is circle and find its centre and radius if $\frac{z-i}{z-1}$ is purely imaginary.

Answer 1

Let $z=x+iy$

$\frac{z-i}{z-1}=\frac{x+iy-i}{x+iy-1}=\frac{x+i(y-1)}{(x-1)+iy}=\frac{x+i(y-1)}{(x-1)+iy}\times \frac{(x-1)-iy}{(x-1)-iy}=\frac{x(x-1)-ixy+(y-1)(x-1)i-i^{2}y}{{(x-1)}^2-{\left(iy\right)}^2}=\frac{x^2-x+y^2+i(xy-x-y+1-xy)}{{(x-1)}^2+{\left(y\right)}^2}=\frac{x^2-x+y^2+i(1-x-y)}{{(x-1)}^2+{\left(y\right)}^2}=\frac{x^2-x+y^2}{{(x-1)}^2+{\left(y\right)}^2}+\frac{(1-x-y)}{{(x-1)}^2+{\left(y\right)}^2}i$

If number is purely imaginary then its real part is zero

$\Rightarrow \frac{x^2-x+y^2}{{(x-1)}^2+{\left(y\right)}^2}=0\Rightarrow x^2-x+y^2=0\Rightarrow x^2+y^2-x=0$

whcih represents a circle

Hence proved