Question

If $z$ is a complex number satisfying the equation $\left(\left|z - \left(1 + i\right)\right|\right)^{2}=2$ and $\omega =\frac{2}{z} \, \left(z \neq 0\right),$ 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 $z=x+iy$ )
$\Rightarrow x^{2}+y^{2}=2\left(x + y\right)\ldots \ldots ..\left(i\right)$
Let, $\omega =h+ik=\frac{2}{z}=\frac{2}{x + i y}=\frac{2 \left(x - i y\right)}{x^{2} + y^{2}},$ so
$h=\frac{2 x}{x^{2} + y^{2}},k=\frac{- 2 y}{x^{2} + y^{2}}$
$\Rightarrow h-k=\frac{2 \left(x + y\right)}{x^{2} + y^{2}}=1$ (from equation $\left(i\right)$ )
$\therefore $ Locus of the point $\omega \left(h , k\right)$ will be $x-y=1.$