Question

If the point P denotes the complex number $z=x+iy$ in the Argand plane and if $\frac{z-i}{z-1}$ is purely imaginary number, find the locus of P.

Answer 1

$z=x+iy\left(\text{Given}\right)$

Therefore,

$\frac{z-i}{z-1}$

$=\frac{(x+iy)-i}{(x+iy)-1}$

$=\frac{x+(y-1)i}{(x-1)+yi}$

$=\frac{x+(y-1)i}{(x-1)+yi}\times \frac{(x-1)-yi}{(x-1)-yi}$

$=\frac{x(x-1)+y(y-1)+i(y-1)(x-1)-xyi}{{(x-1)}^2+y^2}$

$=\frac{(x^2+y^2-x-y)+i(1-x-y)}{{(x-1)}^2+y^2}$

$=\frac{(x^2+y^2-x-y)}{{(x-1)}^2+y^2}+i\frac{(1-x-y)}{{(x-1)}^2+y^2}$

Given that $\frac{z-i}{z-1}$ is purely imaginary.

Therefore,

$Re\left(\frac{z-i}{z-1}\right)=0$

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

$x^2+y^2-x-y=0$

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

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

Hence the locus of $P$ is a circle of radius $1$ having centre at $(\frac{1}{2},\frac{1}{2})$.