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

Solution :
$$\because \,\,{\text{Im}}\left( z \right) \ne 0$$
⇒ $$z$$ is non real and equation $${z^2} + z + \left( {1 - a} \right) = 0$$     will have non real roots, if $$D < 0$$
$$\eqalign{ & \Rightarrow \,\,1 - 4\left( {1 - a} \right) < 0 \cr & \Rightarrow \,\,4a < 3 \cr & \Rightarrow \,\,a < \frac{3}{4} \cr} $$
∴ $$a$$ can not take the value $$\frac{3}{4}$$