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

Answer 1

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

As, $a$ is real,
so $a=\overline{a}$ gives
$\Rightarrow z^2+z+1={\overline{z}}^2+\overline{z}+1$
$\Rightarrow (z-\overline{z})(z+\overline{z}+1)=0$
As, $z\ne \overline{z}$
So, $z+\overline{z}=-1$
$\Rightarrow x=-\frac{1}{2}$ [where $z=x+iy$]
Now,
$a=z^2+z+1$
$={(-\frac{1}{2+iy})}^2+(-\frac{1}{2}+iy)+1$
$=\frac{3}{4}-y^2$
As, $y\ne 0$ so, $a\ne \frac{3}{4}$