Question

If $z$ is a Complex number satisfying the equation ${\left|z-\left(1+i\right)\right|}^2=2$ and $\omega =\frac{2}{z},$ then the locus traced by $\omega$ in the complex plane is

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

Answer 1

The correct option is B $x+y-1=0$

Given:

$\omega =\frac{2}{z}$

$z=\frac{2}{\omega }$

$z-(1+i)=\frac{2}{\omega }-(1+i)$

${|z-(1+i)|}^2={|\frac{2}{\omega }-(1+i)|}^2=2$

Let $\omega =x+iy$

$\Rightarrow {|\frac{2}{x+iy}-(1+i)|}^2=2$

$\Rightarrow \left|\frac{2(x-iy)-(1+i)(x^2+y^2)}{x^2+y^2}\right|=\sqrt{2}$

$\Rightarrow \left|\frac{(2x-x^2-y^2)-i(2y-x^2-y^2)}{x^2+y^2}\right|=\sqrt{2}$

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

$\Rightarrow 4x^2-4x(x^2+y^2)-4y(x^2+y^2)=-4y^2$

$\Rightarrow 4(x^2+y^2)=4(x^2+y^2)(x+y)$

$\Rightarrow x+y=1\to x+y-1=0$