Question

If $arg\left(\frac{z-\omega }{z-{\omega }^2}\right)=0$ then prove that $Re\left(z\right)=-\frac{1}{2}$ ( $\omega$ and ${\omega }^2$ are non-real cube roots of unity).

Answer 1

Remember that if arg (number)$=0$

then number must be a real number

i.e Arg $\left(\frac{z-w}{z-w^2}\right)=0\Rightarrow \frac{z-w}{z-w^2}=m\left(R\right).....\left(1\right)$

also $w=\frac{-1+\sqrt{3}i}{2}{w}^2=\frac{-1-\sqrt{3}i}{2}....\left(2\right)$

from $\left(1\right)$

$\Rightarrow z-w=mz-w^{2}m$

$\Rightarrow z(1-m)=w-w^{2}m$

taking real parts

$Re\left(z\right)(1-m)=Rew-Re{w}^{2}m$

from $\left(2\right)$

$Re\left(z\right)(1-m)=\frac{-1}{2}-\left(\frac{-1}{2}\right)m$

$Re\left(z\right)(1-m)=\frac{-1}{2}(1-m)$

or, $Re\left(z\right)=\frac{-1}{2}$