Let's write $\alpha$ and $\beta$ in the form $a+bi$. $\beta$ is the conjugate of $\alpha$, so if $\alpha=a+bi$, then $\beta=a-bi$. Right. Now we can use these expressions to solve the problem. We know that $\frac{\alpha}{\beta^2}$ is real, so $\frac{a+bi}{(a-bi)^2}$ is real. That would give us $\frac{{(a+bi)^3}}{(a^2+b^2)^2}$. So the imaginary part in the numerator is $3a^2bi-b^3i$. That must be equal to 0. That means $b=0$ or $b=\pm 3a$. We also know that $|\alpha-\beta|=2\sqrt{3}$. So $|2bi|=2\sqrt{3}$. That means $|b|=\sqrt{3}$. So now we have $b=0$ or $b=\pm \sqrt3a$. Right. So if $b=0$, then $|b|=\sqrt{3}$, which is a contradiction. So we must have $b=\pm \sqrt3a$. And since our job is to find $|\alpha|$, we can just find $|a+bi|$ Which is $\sqrt{a^2+b^2}$. We know that $b=\sqrt3a$, so $\sqrt{a^2+b^2}=\sqrt{a^2+3a^2}$. So $|\alpha|=\sqrt{4a^2}=2|a|$. But we also know that $|b|=\sqrt3$. So $|\alpha|=2|a|=2$.


Let $\alpha$ and $\beta$ be conjugate complex numbers such that $\frac{\alpha}{\beta^2}$ is a real number and $|\alpha - \beta| = 2 \sqrt{3}.$  Find $|\alpha|.$