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

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

• Where have you found the problem? What are your thoughts?
– user
May 18 '20 at 7:00
• I have added the picture where it is given position of $z_1,z_2,z_3$ we need to find $\frac{z_b-z_a}{z_c-z_a}$ but not able to find the angle.$P=z_1,Q=z_2,R=z_3$ May 18 '20 at 7:05
• $\frac{z_3-z_1}{z_2-z_1}$ being imaginary implies that the points $z_2$ and $z_3$ are diametrically opposite. Drawing the figure like that and some constructions will give $A$ as 45 degrees. May 18 '20 at 7:47

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 $$\angle B'A'C'=\frac\pi2.$$
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 $$\angle BAC = \frac\pi4$$.