If in a complex plane $z_1$ is represented by point $A, z_2$ by $B$ and $z_3$ by $C$, then $z_3-z_2$ represents the vector $\vec {BC}$, and $z_2-z_1$ represents vector $\vec {AB}$. Thus, these points $A,B,C$ are collinear iff $\vec{AB}$ and $\vec{BC}$ are parallel as they already have one point $B$ in common $\implies \vec {BC}=c\vec{AB}\implies z_3-z_2=c(z_2-z_1)$.
Here, $c$ is taken real because when we multiply by a complex no. , it's real part gives the scaling and imaginary part rotates it. Here, we need the vectors to be parallel ,so we omit the rotation and $c$ has to be real only.
First, let's get a geometric interpretation of the statement "the angles of $\alpha$ and $\beta$ are the same".
That's a picture of $z_1$, $z_2$, and $z_3$. Note that $\theta_1$ and $\theta_2$ represent the angles of $z_3 - z_1$ and $z_3 - z_2$. Therefore, the angle of $\alpha$ (which is $\theta_1 - \theta_2$) is the measure of the geometric angle $z_1z_3z_2$.
So rephrasing the problem in geometric terms, we have:
Suppose four points $A$, $B$, $C$, and $D$ are in the plane such that the angles $BCA$ and $ADB$ are equal. Prove that all four points are in a straight line or lie on a common circle. Conversely, given four points $A$, $B$, $C$, and $D$ either on a line or all on a circle, prove that the angles $ACB$ and $ADB$ are equal.
Where we add the restriction that's unusual for geometry - but is inherited from the complex plane - that an angle $XYZ$ is measured counter-clockwise from $YZ$ to $YX$. (so that the measure of the angle $XYZ$ is $2\pi$ radians minus the measure of the angle $ZYX$) We also allow two angles to be considered equal if their measures differ by exactly $\pi$ radians, to reflect the idea that $z$ and $-z$ have the same angle.
First let's tackle the direction "on a line or a circle implies equal angles".
Now, if the four points lie on a line then obviously the angles $BCA=ACB$ and $ADB$ are equal (since they're both $0$ radians or $\pi$ radians).
If the four points lie on a circle, then:
This shows $A$, $B$, and $C$ all on a circle, and two different possibilities for point $D$: either it lies on the same side of the line $\overline{AB}$ as point $C$ (case $D_2$) or it lies on the opposite side. (case $D_1$)
By the central angle theorem, $\theta_1 = {1 \over 2} \phi = \theta_2$, so if $D$ is in case $D_2$ we have the angles equal. If $D$ is in case $D_2$, we have that
\begin{align}
2\pi - \theta_3 &= {1 \over 2} (2\pi - \phi) \\
&= \pi - {1 \over 2} \phi \\
&= \pi - \theta_1
\end{align}
And therefore $\theta_3 = \theta_1 + \pi$. (which we were going to consider equal)
For the other direction, first take the case that both angles are $0$ or $\pi$. If that's the case, then $C$ and $D$ must both line on line $\overline{AB}$ and we're done. (this corresponds to $\alpha$ and $\beta$ both being real numbers)
Now, if we have some other angle value, then $A$, $B$, and $C$ do not all lie on a line, and therefore there exists a unique circle $O_1$ containing $A$, $B$, and $C$. Likewise, there is a unique circle containing points $A$, $B$, and $D$. Since the central angle $AO_1B$ is equal to the central angle $AO_2B$ (both are twice the angle $ACB$), the radii of the two circles are identical, and from there it's just a bit of case analysis (taking the case where $C$ and $D$ are on the same side of $\overline{AB}$ or on opposite sides, and whether the measures of $ACB$ and $ADB$ are really equal or are off by $\pi$ radians) to show that both circles are identical.
Best Answer
Basically it boils down to showing that:
Proof: Let $P (z) $ be any point on $\bar \alpha z+\bar z \alpha=r $ and let $A $ and $B $ represent $z_1$ and $z_2$. Then we have $AP=BP \Rightarrow |z-z_1|^2=|z-z_2|^2 \Rightarrow (\bar z_2 -\bar z_1)z +(z_2-z_1)\bar z=z_2\bar z_2- z _1\bar z_1$. This is equivalent to the line $\bar \alpha z +\bar z \alpha=r$
Therfore $$\frac {\bar z_2- \bar z_1}{\bar \alpha}=\frac {z_2-z_1}{\alpha}=\frac {z_2\bar z_2 -z_1 \bar z_1}{c} = k $$ Now finding $\bar z_1\alpha +z_2\bar \alpha $ gives us $r $ proving the result to be true. Hope it helps.