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Q.
$z_{1}, z_{2}$ are two non-real complex numbers such that $\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}=1$. Then $z_{1}, z_{2}$ and the origin
Complex Numbers and Quadratic Equations
Solution:
If $\frac{z_{1}}{z_{2}}=z$, the given equation becomes
$z^{2}-z+1=0$
i.e., $z=-\omega$ and $-\omega^{2}$
or, $ \frac{z_{1}}{z_{2}}=-\omega $
$\Rightarrow z_{1}=-z_{2} \omega$
$O B=\left|z_{2}-0\right|=\left|z_{2}\right|$
$O A=\left|z_{1}-0\right|=\left|-z_{2} \omega\right|=\left|z_{2}\right||-\omega|=\left|z_{2}\right|$
and $ A B \left|z_{2}-z_{1}\right|=\left|z_{2}+z_{2} \omega\right| $
$=\left|z_{2}\right||1+\omega|=\left|z^{2}\right|\left|-\omega^{2}\right|=\left|z_{2}\right| $
Thus $z_{1}, z_{2}$ and origin form an equilateral triangle.