# complex number with geometry 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.

• What is an imaginary cube root of unity? – José Carlos Santos Nov 28 '19 at 13:05
• @JoséCarlosSantos Good catch; I read right past that. – saulspatz Nov 28 '19 at 14:01
• 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. – Arthur Nov 28 '19 at 14:54
• @Arthur its a multiple choice question so that ain't bad – Augusta ASAKA Nov 28 '19 at 15:22
• My point is, (A) and (C) are the same choice. As are (B) and (D). That is bad. – Arthur Nov 28 '19 at 15:32

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.

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

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.

• 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 – Augusta ASAKA Nov 28 '19 at 15:27