Complex plane region for complex numebers satisfying $Im\left(\frac{z-z_1}{z-z_2}\right)=0$ and $Re\left(\frac{z-z_1}{z-z_2}\right)=0$

Question

I am being asked to find and plot the geometric region on the complex plane for the following sets: $$ \left\{ z \middle| Im\left(\frac{z-z_1}{z-z_2}\right)=0\right\} $$

$$ \left\{ z \middle| Re\left(\frac{z-z_1}{z-z_2}\right)=0\right\} $$

I am not sure on how to begin the solution.

Answer 1

Consider points $z,z_{1},z_{2}$ on a complex plane.

$Im\left(\frac{z-z_{1}}{z-z_{2}}\right)=0$ if and only if $\angle z_{1}zz_{2}=0,\pi$ i.e. $z$ lies on the line that pass through $z_{1}$ and $z_2$.

$Re\left(\frac{z-z_{1}}{z-z_{2}}\right)=0$ if and only if $\angle z_{1}zz_{2}=\pm\frac{\pi}{2}$ i.e. $z$ lies on the circle with diameter $|z_{1}-z_{2}|$ centered at $\frac{z_{1}+z_{2}}{2}$

Answer 2

1) $ Im\left(\frac{z-z_1}{z-z_2}\right)=0$ leads to,

$$\frac{z-z_1}{z-z_2}-\frac{\bar z-\bar z_1}{\bar z-\bar z_2}=0\implies (z-z_1)(\bar z-\bar z_2) =(z-z_2)(\bar z-\bar z_1)\implies$$ $$(z_2-z_1)\bar z - (\bar z_2 -\bar z_1)z =z_2\bar z_1-z_1\bar z_2$$

which is a linear function in $z$. Note that $z=z_1$ and $z=z_2$ satisfy the equation. Thus, $z$ represents a line in the complex plane that passes the points $z_1$ and $z_2$.

2) $Re\left(\frac{z-z_1}{z-z_2}\right)=0$ leads to, $$\frac{z-z_1}{z-z_2}+\frac{\bar z-\bar z_1}{\bar z-\bar z_2}=0\implies (z-z_1)(\bar z-\bar z_2) +(z-z_2)(\bar z-\bar z_1)=0\implies$$

$$|z|^2-\frac{z_1+z_2}2\bar z-\frac{\bar z_1+\bar z_2}2z + z_2\bar z_1+z_1\bar z_2=0\implies $$ $$\bigg|z-\frac{z_1+z_2}2\bigg|^2=\bigg|\frac{z_1-z_2}2\bigg|^2$$

which represents a circle of the center $\frac{z_1+z_2}2$ and the radius $|\frac{z_1-z_2}2|$.