Question

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 and $Re\left(e^{i\theta }\right)>Re\left(\omega \right)$, 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 $O$)
$\Rightarrow z^2=\omega \overline{\omega }=1$
$\Rightarrow z=-1$
($\because A$ and $D$ are on opposite sides of $BC$)