If $z_1+z_2\cos \alpha+z_3\sin \alpha=0$ then Find value of $\bar z_2z_3+z_2\bar z_3$ If $A(z_1),B(z_2)$ and $C(z_3)$ be the vertices of a $\triangle ABC$
such that $|z_1|=|z_2|=|z_3|=1$
and there exist $\alpha\in \left(0,\frac {\pi}{2}\right)$ such that $z_1+z_2\cos \alpha+z_3\sin \alpha=0$
then
Find value of $\bar z_2z_3+z_2\bar z_3$
My Attempt
I managed to get the answer using a purely algebraic approach. But I feel there is a solution by  geometrical  approach lurking nearby.
$\left|z_2\cos \alpha+z_3\sin \alpha\right|=|-z_1|=1$
$\left|z_2\cos \alpha+z_3\sin \alpha\right|^2=1$
$(z_2\cos \alpha+z_3\sin \alpha)(\bar z_2\cos \alpha+\bar z_3\sin \alpha)=1$
$|z_2|^2\cos^2\alpha+|z_3|^2\sin^2\alpha+(z_2\bar z_3+\bar z_2z_3)\cos \alpha\sin \alpha=1$
$\cos^2\alpha+\sin^2\alpha+(z_2\bar z_3+\bar z_2z_3)\cos \alpha\sin \alpha=1$
$(z_2\bar z_3+\bar z_2z_3)\cos \alpha\sin \alpha=0$
$z_2\bar z_3+\bar z_2z_3=0$
Here I wonder if there is some other approach to reach this conclusion
 A: We can represent your question in the complex world to ease our calculations in geometry:
Let $z_1, z_2, z_3$ be complex values (vectors in the complex plane) that have a length of $1$.
This means $z_1, z_2, z_3$, as points, must lie on the radial equation $x^2 + y^2 = 1$. Visually, it would look similar to this:

All complex vectors can be represented in the polar form $re^{i\theta}, i^2 = -1$ ($r$ is the length, $1$, in this context) ;
$z_1 = e^{i\theta_1}, z_2 = e^{i\theta_2}, z_3 = e^{i\theta_3}$
A conjugate of a complex number, $\overline{z}$, is equivalent to $e^{-i\theta}$, if $z = e^{i\theta}$
Hence, $\overline{z_1} = e^{-i\theta_1}, \overline{z_2} = e^{-i\theta_2}, \overline{z_3} = e^{-i\theta_3}$
Now, notice the particular angle geometry between the opposite conjugate angles added...

They have the same angle difference, in the reverse direction!
That is, $\theta_{2} + (-\theta_{3}) = -(\theta_{3} + (-\theta_{2}))$
By index laws (and De Movire's theorem), $z_2\overline{z_3} = e^{i\big(\theta_{2} + (-\theta_{3})\big)}$
Hence,
$$\theta_{2} + (-\theta_{3}) = -(\theta_{3} + (-\theta_{2})) \implies z_2\overline{z_3} = -z_3\overline{z_2} \to z_2\overline{z_3}  + z_3\overline{z_2} = 0$$
