Question

If $z$ is a complex number satisfying the equation ${\left(\left|z-\left(1+i\right)\right|\right)}^2=2$ and $\omega =\frac{2}{z}\left(z\ne 0\right),$ then the locus of $\omega$ is

• A. $x-y-1=0$
• B. $x+y-1=0$
• C. $x-y+z=0$
• D. $x+2y+1=0$

Answer 1

Answer: (A)

We have, ${\left(\left|z-\left(1+i\right)\right|\right)}^2=2$
$\Rightarrow {\left(x-1\right)}^2+{\left(y-1\right)}^2=2$ (Putting $z=x+iy$ )
$\Rightarrow x^2+y^2=2\left(x+y\right)\dots \dots ..\left(i\right)$
Let, $\omega =h+ik=\frac{2}{z}=\frac{2}{x+iy}=\frac{2\left(x-iy\right)}{x^2+y^2},$ so
$h=\frac{2x}{x^2+y^2},k=\frac{-2y}{x^2+y^2}$
$\Rightarrow h-k=\frac{2\left(x+y\right)}{x^2+y^2}=1$ (from equation $\left(i\right)$ )
$\therefore$ Locus of the point $\omega \left(h,k\right)$ will be $x-y=1.$