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Q.
Let $z$ be non real number such that $\frac{1+z+z^2}{1-z+z^2} \in R$, then value of $7|z|$ is
Complex Numbers and Quadratic Equations
Solution:
$\frac{1+z+z^2}{1-z+z^2}=1+\frac{2 z}{1-z+z^2}$ is real
$ \Rightarrow 1+\frac{2}{z+\frac{1}{z}-1}$ is real
$\Rightarrow z+\frac{1}{z}$ is real
$\Rightarrow z+\frac{1}{z}=\bar{z}+\frac{1}{\bar{z}} $
$ \Rightarrow (z-\bar{z})=\frac{1}{\bar{z}}-\frac{1}{\bar{z}}$
$\Rightarrow (z-\bar{z})\left(1-\frac{1}{|z|^2}\right)=0 $
$ \Rightarrow |z|=1$