### Instruction: 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 $|\alpha|.$.### Response: Since $\alpha$ and $\beta$ are conjugate complex numbers, I can write them as $\alpha = a + bi$ and $\beta = a - bi$ for some real numbers $a$ and $b$.
Then, $\frac{\alpha}{\beta^2} = \frac{a + bi}{(a - bi)^2} = \frac{(a + bi)(a + bi)}{(a - bi)(a + bi)} = \frac{a^2 + 2abi - b^2}{a^2 + b^2}.$
This is a real number if and only if the imaginary part is zero, that is, $2ab = 0.$
This means that either $a = 0$ or $b = 0.$
If $a = 0,$ then $\alpha = bi$ and $\beta = -bi,$ so $|\alpha - \beta| = |2bi| = 2|b|.$
But I know that $|\alpha - \beta| = 2 \sqrt{3},$ so this implies that $|b| = \sqrt{3}.$
Therefore, $|\alpha| = |bi| = |b| = \sqrt{3}.$
If $b = 0,$ then $\alpha = a$ and $\beta = a,$ so $|\alpha - \beta| = |0| = 0.$
But this contradicts the fact that $|\alpha - \beta| = 2 \sqrt{3},$ so this case is impossible.
Hence, the only possibility is that $a = 0$ and $b = \sqrt{3},$ and the answer is $|\alpha| = \sqrt{3}.$