What data is the table referring to?
Answer:

Step-by-step explanation:
Given that alpha and beta be conjugate complex numbers
such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.
Let

since they are conjugates


Imaginary part of the above =0
i.e. 
So the value of alpha = 
1/25 i don’t know if this is right but here you go