Question

If the complex number associated with the vertices $A,B,C$ of $\Delta ABC$ are $e^{i\theta },\omega ,\overline{\omega }$, respectively [Where $\omega ,\overline{\omega }$ are the complex cube roots of unity of $cos\theta >Re\left(\omega \right)]$, then the complex number of the point where angle bisector of $A$ meets the circumcircle of the triangle is

• A. $e^{i\theta }$
• B. $e^{-i\theta }$
• C. $\omega ,\overline{\omega }$
• D. $\omega +\overline{\omega }$

Answer 1

The correct option is D $\omega +\overline{\omega }$

Clearly,
$\angle DOB=\angle COD=A$
$\Rightarrow z=\omega e^{iA}$ and $\overline{\omega }=z{e}^{iA}$
(Applying rotation about O)
$\Rightarrow z^2=\omega \overline{\omega }=1$
$\Rightarrow z=-1$
(As A and D are one opposite sides of BC)