Mathematics - Complex Number Question with Solution | TestHub

Given $\overline{z}\left(1-i\right)=z^2\left(1+i\right)$

So $\left|\overline{z}\right|\left|1-i\right|={\left|z\right|}^2\left|1+i\right|$

$\Rightarrow \left|z\right|={\left|z\right|}^2\Rightarrow \left|z\right|=0$ or $\left|z\right|=1$

Also, $\overline{z}\left(1-i\right)=z^2\left(1+i\right)\Rightarrow \overline{z}\frac{\left(1-i\right)}{\left(1+i\right)}=z^2$

$\Rightarrow -i\overline{z}=z^2$

Let $z=x+iy$

$\Rightarrow x^2-y^2-\left(2xy\right)i=y-ix$

i.e. $x^2-y^2=y$ and $-2xy=-x$

From $-2xy=-x$, we get

$x=0 \text{or} y=-\frac{1}{2}$

i.e. the points are $\left(0,0\right)$, $\left(0,1\right)$, $\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$ and $\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$