Question

If $\omega$ is any complex number such that $z\omega =|z{|}^2$ and $|z-\overline{z}|+|\omega +\overline{\omega }|=4$, then as $\omega$ varies, then the area bounded by the locus of $z$ is

• A. $4$ sq. units
• B. $8$ sq. units
• C. $16$ sq. units
• D. $12$ sq. units

Answer 1

Answer: (B)

$8$ sq. units

$z\omega ={\left|z\right|}^2$

$z\omega =z\overline{z}\Rightarrow \omega =\overline{z}$

$|z-\overline{z}|+|\omega +\overline{\omega }|=|z-\overline{z}|+|z+\overline{z}|=4$

Put $z=x+iy$

$\left|(x+iy-x+iy)\right|+|x+iy+x-iy|=4$

$\left|2iy\right|+\left|2x\right|=4$

$\left|2y\right|+\left|2x\right|=4$

$\left|x\right|+\left|y\right|=2$

Now, $\left|x\right|+\left|y\right|=2$ has graph of square with side$=\frac{4}{\sqrt{2}}$

Therefore area$={\left(\frac{4}{\sqrt{2}}\right)}^2=\frac{16}{2}=8$ square units