ABC is an acute-angled triangle whose altitude from the vertices meet the circumcircle

Question

Let ABC be an acute-angled triangle. Let the altitudes from the vertices A, B, C meet the circumcircle at P, Q, R whose corresponding complex numbers are $z_1,z_2$ and $z_3$ respectively. If is $\frac{z_3-z_1}{z_2-z_1}$ is imaginary number then find the value of angle A.

My approach is illustrated below but not able to approach

Answer 1

Let $A',B',C'$ be the intersection points of the altitudes with the circumscribed circle. Then we have: $$ \angle BAC=\frac{\pi-\angle B'A'C'}2. $$ From $$\Re\frac{z_3-z_1}{z_2-z_1}=0$$ we know $$\angle B'A'C'=\frac\pi2.$$

Can you take it from here?

Answer 2

From the given, we have $\frac{z_3-z_1}{z_2-z_1} = e^{i\frac\pi2}$, or $Arg \left( \frac{z_3-z_1}{z_2-z_1}\right) =\frac\pi2$

\begin{align} \angle BAC & = \angle BAP + \angle CAP \\ & = \angle BQP + \angle CRP \\ & = Arg \left( \frac{z_2-z_1}{z_2-b} \right) + Arg \left( \frac{z_3-c}{z_3-z_1} \right) \\ & = Arg \left( \frac{z_2-z_1}{z_3-z_1} \frac{z_3-c}{z_2-b} \right) \\ & = Arg \left( \frac{z_2-z_1}{z_3-z_1}\right) +Arg\left( \frac{z_3-c}{z_2-b} \right) \\ & = -\frac\pi2 +\angle RXQ \\ & = - \frac\pi2 +(\pi - \angle BAC) \\ \end{align}

which yields $\angle BAC = \frac\pi4$.