I am trying to solve a PDE by applying separation of variables using the complex form of a Fourier Series. The PDE to solve is $$\bigtriangleup u = 0,\ u(1,\ \theta) = 2\pi\theta - \theta^2$$
I plugged $u = R(r)T(\theta)$ into the polar Laplacian formula, and ended up with $\frac{T_{\theta \theta}}{T} = \lambda$ and $-\frac{r^2 R_{rr} + r R_r}{R} = \lambda$. My solutions to the Sturm-Liouville problems are $R = Ar^{-\sqrt{-\lambda}} + Br^{\sqrt{-\lambda}}$ and $T = Ce^{-\sqrt\lambda \theta} + De^{\sqrt\lambda \theta}$ (I have ruled out the $\lambda = 0$ case). Thus I have $$u = (Ar^{-\sqrt{-\lambda}} + Br^{\sqrt{-\lambda}})(Ce^{-\sqrt\lambda \theta} + De^{\sqrt\lambda \theta})$$
Since the given boundary condition implies that the domain is a disk, I also have the implied boundary conditions $u(r,\ 0) = u(r,\ 2\pi)$ and $u_\theta (r,\ 0) = u_\theta (r,\ 2\pi)$. Plugging these implied conditions into the general solution yields an algebraic system with solution $\lambda = -n^2$, where $n$ is an arbitrary integer, therefore $$u = \sum_{n = -\infty}^\infty (A_n r^{-n} + B_n r^n)(C_n e^{-n \theta i} + D_n e^{n \theta i})$$
Now I think the RHS expression here has a redundancy which allows me to kill $C_n e^{-n \theta i}$ without losing any of the correct terms, because of the two-sided summation. Then I can lump $D_n$ into $A_n$ and $B_n$ simplifying things to $$u = \sum_{n = -\infty}^\infty (A_n r^{-n} + B_n r^n)e^{n \theta i}$$
Applying the given boundary condition yields $2\pi\theta - \theta^2 = \sum_{n = -\infty}^\infty (A_n + B_n)e^{n \theta i}$. The Fourier coefficient $A_n + B_n$ should be given by $\frac{1}{2\pi} \int_0^{2\pi} (2\pi\theta - \theta^2)e^{-n\theta i} d\theta$, which is $-\frac{2}{n^2}$. This leads to $u = \sum_{n = -\infty}^\infty (A_n r^{-n} + (-\frac{2}{n^2} - A_n)r^n)e^{n \theta i}$, which expands to $$u = \sum_{n = -\infty}^\infty \frac{(A_n n^2 r^{-n} - A_n n^2 r^n - 2r^n)(cos(n\theta) + sin(n\theta)i)}{n^2}$$
I think this answer is incorrect, because if we consider the special case of the problem which models an equilibrium temperature distribution, then $A_n$ must be $0$ (because $u$ must be finite at the origin), and $u$ must be real, but in my answer $u$ is not in fact real when $A_n = 0$. What am I doing incorrectly to arrive at this solution?