Question

If $z=x+iy,|z|=1$ and $\omega =\frac{(1-z{)}^2}{1-z^2}$, then the locus of $\omega$ is equivalent to

• A. $|z-2|=|z+2|$
• B. $|z-2i|=|z+2i|$
• C. $|z-1-i|=|z+1-i|$
• D. $|z-i|=|z+i|$

Answer 1

Answer: (A), (C)

The correct options are
A $|z-2|=|z+2|$
C $|z-1-i|=|z+1-i|$

Given : $\omega =\frac{(1-z{)}^2}{1-z^2}$
$\Rightarrow \omega =\frac{1-z}{1+z}=\frac{z\overline{z}-z}{z\overline{z}+z}\Rightarrow \omega =\frac{\overline{z}-1}{\overline{z}+1}=-\overline{\omega }\Rightarrow \omega +\overline{\omega }=0$
So, $\omega$ should be purely imaginary

Hence, the locus of $\omega$ is $x=0$
Now, by obersvation, we see that all the options given represents perpendicular bisector.
$|z-2|=|z+2|$
Perpendicular bisector of points $(2,0)$ and $(-2,0)$
Which is $x=0$

$|z-2i|=|z+2i|$
Perpendicular bisector of points $(0,2)$ and $(0,-2)$
Which is $y=0$

$|z-1-i|=|z+1-i|\Rightarrow |z-(1+i)|=|z-(-1+i)|$
Perpendicular bisector of points $(1,1)$ and $(-1,1)$
Which is $x=0$

$|z-i|=|z+i|$
Perpendicular bisector of points $(0,1)$ and $(0,-1)$
Which is $y=0$