In this question, the comment suggests that the imaginary terms in the stated solution might sum to $0$. In order for this to happen, it must be the case that
$\sum_{-\infty}^\infty [\frac{2(-1)^{n + 1} r^n \sin(n\theta) + 2(-1)^n n\pi r^n \cos(n\theta)}{n^2} + Im(\frac{A_n n^2 r^{-n} \cos(n\theta) - A_n n^2 r^n \cos(n\theta) + A_n n^2 r^{-n} \sin(n\theta)i - A_n n^2 r^n \sin(n\theta)i}{n^2})i] = 0$
This should hold for all real numbers $r$ and $\theta$, but if it does not hold for all $A_n$ (which can be complex), finding the $A_n$'s where it does work could help narrow down the solution. It should at least hold for $A_n = 0$ if the PDE solution is to have a physical interpretation of equilibrium temperature.
I recall from solving ODEs via power series that it is often legal to drop a summation operator in an equation like this and assume that each term must equal $0$ in and of itself, but I am not sure if this works with the complex numbers here, nor the two-sided infinite summation. I also do not know how to deal with the imaginary part operator which emerges due to the uncertainty of whether the $A_n$'s are real or imaginary.
How can I see if this equation if true, or if not, for which value of $A_n$ it is true?