Question

Let $z$ be non real numbers such that $\frac{1+z+z^2}{1-z+z^2}ϵR$, then value of $7\mid z\mid$ is

• A. $1$
• B. $3$
• C. $5$
• D. $7$

Answer 1

Answer: (D)

$7$

$\frac{z^2+z+1}{z^2-z+1}=1+\frac{2z}{z^2-z+1}$

So the given fraction is real if and only if the fraction $\frac{z}{z^2-z+1}$ is real. But a fraction is real if and only if its reciprocal is, so we need:

$\frac{z^2-z+1}{z}=z-1+z^{-1}$

To be a real number. So we get:

The fraction is real if and only if $z+z^{-1}$ is real (or $z=0$).

Now if $z=a+bi$, with $z\ne 0$, the imaginary part of $z+z^{-1}$ is $b-\frac{b}{a^2+b^2}=\frac{b(a^2+b^2-1)}{a^2+b^2}=\frac{b(|z|-1)}{|z|}=0$. Hence we get:

Either $|z|=1$ or $z$ is real.

Hence $7|z|=7\times 1=7$.