Given, $$\quad a=z^2+z+1$$

$$\Rightarrow z^2+z+1-a=0$$

Using Sridharacharya Rule

$$\begin{aligned} & \qquad z=\frac{-1 \pm \sqrt{1^2-4(1-a)}}{2} \\ & \Rightarrow \quad z=\frac{-1 \pm \sqrt{1-4+4 a}}{2} \\ & \Rightarrow \quad z=\frac{-1 \pm \sqrt{4 a-3}}{2} \end{aligned}$$

Now, we can observe that if $$\mathrm{a}=\frac{3}{4}$$, then $$z= \frac{-1}{2}$$ which will become purely real.

Hence, if $$a \neq \frac{3}{4}$$, otherwise $$z$$ will be purely real.

(i) Make quadratic equation in $$z$$ and apply Sridharacharya rule.

(ii) Analyse the value of '$$a$$' for which '$$z$$' becomes purely real.