Question

Let z be a complex number such that the imaginary part of z is non-zero 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

$\frac{3}{4}$

The roots of the equation $a=z^2+z+1$ are non-real
$z^2+z+1-a=0\Rightarrow z=\frac{-1\pm \sqrt{\left(4a\right)-3}}{2}$
For z do not have real roots, $4a-3<0\Rightarrow a<\frac{3}{4}$