If $A(z_1)$ and $(z_2)$ are two points in argand plane, find $\angle ABO$ If $A(z_1)$ and $(z_2)$ are two points in argand(complex) plane such that $$\frac{z_1}{z_2}+\frac{\overline{z_1}}{\overline{z_2}}=2$$. Find the value of $\angle ABO$ where $O$ is origin.
Using given condition, I found that Real part of $\frac{z_1}{z_2}=1$ but I am not able to use this to find $\angle ABO$. Could someone help me with this?

If we write $z_1=\rho_1 e^{i\theta_1}$ and $z_2=\rho_2 e^{i\theta_2}$ (with $z_1\ne z_2$, if $z_1=z_2$ we have trivially $\beta=0$), then from the equation we get:
$${\rho_1 e^{i\theta_1}\over\rho_2 e^{i\theta_2}}+{\rho_1 e^{-i\theta_1}\over\rho_2 e^{-i\theta_2}}=2$$
$$e^{i(\theta_1-\theta_2)}+e^{-i(\theta_1-\theta_2)}=2{\rho_2\over\rho_1}$$
$${e^{i(\theta_1-\theta_2)}+e^{-i(\theta_1-\theta_2)}\over2}={\rho_2\over\rho_1}$$
$$\cos(\theta_1-\theta_2)={\rho_2\over\rho_1}$$
so we get $\rho_2\le\rho_1$.
Now we can apply the law of sines to $\triangle ABO$:
$${\sin\beta\over\rho_1}={\sin(\theta_1-\theta_2)\over\overline {AB}}\longrightarrow\sin\beta=\rho_1{\sin(\theta_1-\theta_2)\over\overline {AB}}$$
where $\beta=\angle ABO$.
We know that:
$$\sin(\theta_1-\theta_2)=\sqrt{1-\cos^2(\theta_1-\theta_2)}={\sqrt{\rho_1^2-\rho_2^2}\over\rho_1}$$
and applying the cosine rule to $\triangle ABO$:
$$\overline{AB}=\sqrt{\rho_1^2+\rho_2^2-2\rho_1\rho_2\cos(\theta_1-\theta_2)}=\sqrt{\rho_1^2+\rho_2^2-2\rho_1\rho_2{\rho_2\over\rho_1}}=\sqrt{\rho_1^2-\rho_2^2}$$
hence we get: $\sin\beta=1\longrightarrow\beta={\pi\over2}$. The angle $\beta\equiv\angle ABO$ is always ${\pi\over2}$.
