complex number with geometry

Question

Well i tried this question by putting it in co-ordinate plane from argand plane and
tried the question that way but it got pretty messed up so i gave up on that, btw it is pretty clear that the complex numbers will lie on the circle ; mod(z) = 1 which becomes the circumcirle that way, therefore geometrically i know the ans is -1 but i am having a hard time proving it.

Answer 1

Here's a hint. If the vertices of the triangle are $1$, $e^{i\theta_1}$, and $e^{i\theta_2}$, it's easy to figure out where the angle bisector at $1$ intersects the unit circle. Can you see how to transform the given problem to this easy problem, and then transform the solution back to the original context?

EDIT

I was certain your statement that the answer is always $-1$ was wrong, but it turns out to be correct. My situation is just the reverse of yours. I can prove it with complex numbers, but I don't quite see the geometry.

Answer 2

Here is a geometric proof: Let $P$ be where the angle bisector of $\angle A$ meets the circumcircle (with $P\neq A$, of course). Then, because $\angle BAP=\angle CAP$, those two angles subtend equal length cords of the circumcircle. In other words, $|PB|=|PC|$.

Since $B$ and $C$ are complex conjugates, that means that $P$ is a real number. And there are only two real numbers on the unit circle. $\cos\theta>\text{Re}(\omega)$ lets you decide which one it is.