Given : $\alpha$ and $\beta$ are roots of the equation
$x^2+x+1=0$
Then $x=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2}$
$\therefore \quad \alpha=\frac{-1+\sqrt{3} i}{2}$ and $\beta=\frac{-1-\sqrt{3} i}{2}$
$\alpha=-\frac{1}{2}+i \frac{\sqrt{3}}{2}$ and $\beta=-\frac{1}{2}-\frac{\sqrt{3}}{2} i$
$\Rightarrow \quad \alpha=\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)=\left(e^{i \cdot \frac{2 \pi}{3}}\right)$
$\beta=\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)=\left(e^{i \frac{4 \pi}{3}}\right)$
$\alpha^{2023}=\left(e^{i \frac{2 \pi}{3}}\right)^{2023}=e^{i \times \frac{4046}{3} \pi}$
$=\cos \frac{4046 \pi}{3}+i \sin \left(\frac{4046 \pi}{3}\right)$
$=\cos \left(1348 \pi+\frac{2 \pi}{3}\right)+i \sin \left(1348 \pi+\frac{2 \pi}{3}\right)$
$=\cos \left(\frac{2 \pi}{3}\right)+i \sin \left(\frac{2 \pi}{3}\right) \alpha^{2023}=-\frac{1}{2}+i \frac{\sqrt{3}}{2}=\alpha$
Now,
$\beta^{1012}=\left(e^{i \cdot \frac{4 \pi}{3}}\right)^{1012}=e^{i \cdot \frac{4048}{3}}$
$=\cos \left(\frac{4048 \pi}{3}\right)+i \sin \left(\frac{4048 \pi}{3}\right)$
$=\cos \left(1349 \pi+\frac{\pi}{3}\right)+i \sin \left(1349 \pi+\frac{\pi}{3}\right)$
$=\cos \left(\pi+\frac{\pi}{3}\right)+i \sin \left(\pi+\frac{\pi}{3}\right)=\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}$
$\Rightarrow \quad \beta^{1012}=\beta$
$\because \quad \alpha^{2023}=\alpha$ and $\beta^{1012}=\beta$
$\therefore \quad$ The required equation whose root is $\alpha^{2023}$ and $\beta^{1012}$ is $x^2+x+1=0$.