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enter image description here 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.

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    $\begingroup$ What is an imaginary cube root of unity? $\endgroup$ – José Carlos Santos Nov 28 '19 at 13:05
  • $\begingroup$ @JoséCarlosSantos Good catch; I read right past that. $\endgroup$ – saulspatz Nov 28 '19 at 14:01
  • $\begingroup$ Note that $-1=\omega+\omega^2$ and $1=-(\omega+\omega^2)$ (assuming they mean that $\omega$ is a primitive cube root of unity). I don't know who put together those choices, but they don't seem to have done a very good job in my opinion. $\endgroup$ – Arthur Nov 28 '19 at 14:54
  • $\begingroup$ @Arthur its a multiple choice question so that ain't bad $\endgroup$ – Augusta ASAKA Nov 28 '19 at 15:22
  • $\begingroup$ My point is, (A) and (C) are the same choice. As are (B) and (D). That is bad. $\endgroup$ – Arthur Nov 28 '19 at 15:32
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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.

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  • $\begingroup$ well i am not comfortable with concept equal length cords but if it is true then it sorta answers the question $\endgroup$ – Augusta ASAKA Nov 28 '19 at 15:34
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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?

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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.

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  • $\begingroup$ just make a unit circle in argand plane and mark w and w^2 and any other complex number Z { given Re(z) > -1/2 } and you will see it $\endgroup$ – Augusta ASAKA Nov 28 '19 at 15:27

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