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

$\text{Given: } z \in \mathbb{C},\ \operatorname{Im}(z) \neq 0$

$a = z^{2}+z+1 \in \mathbb{R}$

$\text{To find: which value } a \text{ cannot take}$

$-1,\quad \frac{1}{3},\quad \frac{1}{2},\quad \frac{3}{4}$

$\text{Plan: put } z = x+iy,\ y \neq 0$

$z = x+iy, \qquad y \neq 0$

z^{2} = x^{2}-y^{2}+2ixy

$a = z^{2}+z+1 = (x^{2}-y^{2}+x+1) + i(2xy+y)$

$\operatorname{Im}(a) = 2xy+y = y(2x+1) = 0$

$y \neq 0 \Rightarrow 2x+1 = 0 \Rightarrow x = -\frac{1}{2}$

$a = x^{2}-y^{2}+x+1$

$x = -\frac{1}{2} \Rightarrow a = \frac{1}{4}-y^{2}-\frac{1}{2}+1$

$a = \frac{3}{4}-y^{2}$

$y \neq 0 \Rightarrow y^{2} > 0 \Rightarrow a = \frac{3}{4}-y^{2} < \frac{3}{4}$

$a \in \left(-\infty, \frac{3}{4}\right)$

$a = -1 \Rightarrow y^{2} = \frac{7}{4} > 0 \quad \checkmark$

$a = \frac{1}{3} \Rightarrow y^{2} = \frac{5}{12} > 0 \quad \checkmark$

$a = \frac{1}{2} \Rightarrow y^{2} = \frac{1}{4} > 0 \quad \checkmark$

$a = \frac{3}{4} \Rightarrow y^{2} = 0 \Rightarrow y = 0 \quad \text{✗ not allowed}$

$\boxed{\frac{3}{4}} \quad \text{(option 4)}$