Question

Let z be a complex number such that the imaginary part of z is nonzero and a = $z^2+z+1$ is real. Then a cannot take the value

• A. $-1$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is D $\frac{3}{4}$

Let $z=\alpha +i\beta$. Then

$Im(z^2+z+1)=2\alpha \beta +\beta =0$

$\Rightarrow \alpha =-\frac{1}{2}$ since $\beta \ne 0$. Then

$a={\alpha }^2-{\beta }^2+\alpha +1$

$a=\frac{3}{4}+\beta$

$a\ne \frac{3}{4}$. $[\because \beta \ne 0]$