The problem statement is:
Suppose that the real series $∑_0^{∞} a_n$ and $∑_0^{∞} b_n$ converge absolutely.
Part 1
Prove that there is a function $u(r,θ)$ which is harmonic in $1<r<2$ and continuous onto the boundary such that $$u(1,θ)=∑ancos(nθ)$$
and
$$u(2,θ)=∑bncos(nθ)$$
Part 2
Is $u(r,θ)$ unique?
My work:
We can look at a Laurent series $f(z) = \sum c_nz^n$
$$=\sum c_n r^ne^{in\theta}$$ $$=\sum (a_n + ib_n) r^ne^{in\theta}$$ $$=\sum a_nr^ne^{in\theta} + ib_n r^ne^{in\theta}$$ $$=\sum a_nr^ncos(n\theta)-b_nr^nsin(n\theta)+ i(b_nr^ncos(n\theta)+a_nr^nsin(n\theta))$$
Now I think the goal is to have
$$u(1,\theta) = Re \big( \sum a_nr^ncos(n\theta)-b_nr^nsin(n\theta)+ i(b_nr^ncos(n\theta)+a_nr^nsin(n\theta))\big)$$
$\implies$ $$u(1,\theta) = \sum a_n1^ncos(n\theta)-b_n1^nsin(n\theta)$$
and similarly
$$u(2,\theta) = \sum a_n2^ncos(n\theta)-b_n2^nsin(n\theta)$$
so that, if I can show the R.H.S. is some convergent complex series in the annulus 1<|z|<2, then it defines an analytic function in the annulus, and therefore the L.H.S. of the above two equalities must be harmonic in this annulus, as required.
I am stuck on a few things:
Taking the real part as I had shown above gives an extra term in the summand that I don't want - namely, the $b_nsin(n\theta)$ term. I am guessing that somehow the $b_n$'s must all be zero, but how can I justify this?
Of course, this is one harmonic function, $u(r,\theta)$, from one analytic function on the annulus. So, there needs to be some matching of coefficients of the above two, different, Laurent series. How could I do that?
And, the coefficients have to be chosen, so that the series actually converges in this annulus. Here's where I know I should use the assumptions that the real series $\sum a_n$ and $\sum b_n$ converge absolutely. Any idea how I can do this?
Thanks,
Hint: To match the $a_n \cos (n\theta)$ and $b_n \cos(n\theta)$, take a suitable combination of the real parts of $z^n$ and $z^{-n}$.
EDIT: If this combination is $u_n(r,\theta) = c_n r^n \cos(n \theta) + c_{-n} r^{-n} \cos(n \theta)$, we need $$ \eqalign{c_n + c_{-n} &= a_n \cr 2^n c_n + 2^{-n} c_{-n} &= b_n\cr} $$ Solve this system of equations for $c_n$ and $c_{-n}$. Then show $\sum_{n=-\infty}^{-1} c_n z^n$ converges absolutely and uniformly on $|z| \ge 1$ and $\sum_{n=0}^\infty c_n z^n$ converges absolutely and uniformly on $|z| \le 2$.