Question

If $z=x+iy$ is a complex number satisfying the equation $|z-(1+i){|}^2=2$ and $p=\frac{2}{z}$, then the locus of $p$ in argand plane is

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

Answer 1

The correct option is A $x-y=1$

Let $z=x+iy$
We have $|z-(1+i){|}^2=2$
$\Rightarrow (x-1{)}^2+(y-1{)}^2=2$
$\Rightarrow x^2+y^2=2(x+y)\cdots \left(1\right)$

Let $p=h+ik$, then
$h+ik=\frac{2}{z}\Rightarrow h+ik=\frac{2(x-iy)}{x^2+y^2}\Rightarrow h=\frac{2x}{x^2+y^2},k=-\frac{2y}{x^2+y^2}\Rightarrow h-k=\frac{2(x+y)}{x^2+y^2}$
Using equation $\left(1\right)$, we get
$\Rightarrow h-k=1$

Hence, the locus of $p$ is $x-y=1$