$ a $ can't assume what value?

Question

Let $z$ be a complex number such that the imaginary part of $z$ is not null and $ a = z ^ 2 + z + 1 $ be real. So $ a $ can't assume what value?

I know it may sound easy to you guys, but I can't do it. Thanks in advance.

Answer 1

Let $z = x+\mathrm{i}y$. Then $z^2 + z + 1 = x^2-y^2+x+1 + \mathrm{i}(y+2xy)$. Since $a$ is required real, we have $a = x^2-y^2+x+1$ and $0 = y(1+2x)$.

Since $y = \mathrm{Im}(z) \neq 0$, we must have $x = -1/2$. Then $a = 1/4 - y^2 - 1/2 + 1 = -y^2 + 3/4$. Since $y^2 \geq 0$, this says $a$ cannot assume any value greater than $3/4$.

Answer 2

$a$ is real and we may manipulate the equation to get $$z^2+z+1-a=0\implies z=\frac{-1\pm\sqrt{1-4(1-a)}}2$$ Yet $z$ is not real, so we have $1-4(1-a)<0$ or $a<\frac34$. So $a$ cannot be any real number that is at least $\frac34$.