Question

If $\frac{z-i}{z-1}$ is purely imaginary then the locus of z is

• A. $x^2+y^2+x-y=0$
• B. $x^2+y^2+x+y=0$
• C. $x^2+y^2-2x-3y=0$
• D. $x^2+y^{2}x+3y=0$

Answer 1

The correct option is A $x^2+y^2+x-y=0$

$Let$
$z=x+iy$
$\frac{z-i}{z+1}$
$=\frac{x+iy-i}{x+iy+1}$
$=\frac{x+i(y-1)}{(x+1)+iy}$
$rationali\sin gthedenominator$
$\frac{x+i(y-1)}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy}$
$=\frac{[x+i(y-1)][(x+1)-iy]}{{(x+1)}^2-{\left(iy\right)}^2}$
$=\frac{x(x+1)-xyi+i(y-1)(x+1)-i^{2}y(y-1)}{{(x+1)}^2+y^2}$
$=\frac{x^2+x-xyi+i(xy+y-x-1)+y^2-y}{{(x+1)}^2+y^2}$
$=\frac{(x^2+y^2+x-y)-xyi+xyi+i(y-x-1)}{{(x+1)}^2+y^2}$
$equatingtherealparttozero$
$\frac{x^2+y^2+x-y}{{(x+1)}^2+y^2}=0$
$\Rightarrow x^2+y^2+x-y=0$
$isthelocusofz$