Question

If the real part of $\frac{\overline{z}+2}{\overline{z}-1}$ is $4$, then show that the locus of the point representing z in the complex plane is a circle.

Answer 1

Given : $Re\left(\frac{\overline{z}+2}{\overline{z}-1}\right)=4$

Let $z=x+iy$
$\Rightarrow \overline{z}=x-iy$

Now,
$\frac{\overline{z}+2}{\overline{z}-1}=\frac{x-iy+2}{x-iy-1}$
$\frac{\overline{z}+2}{\overline{z}-1}=\frac{(x+2)-iy}{(x-1)-iy}$
$\frac{\overline{z}+2}{\overline{z}-1}=\frac{(x+2)-iy}{(x-1)-iy}\times \frac{(x-1)+iy}{(x-1)+iy}$
$\frac{\overline{z}+2}{\overline{z}-1}=\frac{(x+2)(x-1)+iy\left[\right(x+2)-(x-1\left)\right]+y^2}{(x-1{)}^2-(iy{)}^2}$
$\{\because i^2=-1\}$
$\frac{\overline{z}+2}{\overline{z}-1}=\frac{(x+2)(x-1)+y^2+i\left(3y\right)}{(x-1{)}^2+y^2}$

Now,
$Re\left(\frac{\overline{z}+2}{\overline{z}-1}\right)=4$
$\Rightarrow \frac{(x+2)(x-1)+y^2}{(x-1{)}^2+y^2}=4$
$\Rightarrow x^2+x-2+y^2=4(x^2+1-2x)+4y^2$
$\Rightarrow x^2+x-2+y^2=4x^2-8x+4+4y^2$
$\Rightarrow 3x^2-9x+6+3y^2=0$
$\therefore x^2+y^2-3x+2=0$

Which is an equation of a circle.
Hence, $Re\left(\frac{z+2}{\overline{z}-1}\right)=4$ represents a circle.