The general solution to the Laplace equation, in polar cooridnates, on a circular domain is given by: $$U(r,\theta) = a_0+\sum_{m=1}^{\infty}r^m(a_m\cos(m\theta) +b_m\sin(m\theta))$$
The coefficients, $a_0$,$a_m$ and $b_m$ can all be found by doing some integrals relating to the boundary condition, $f(\theta) =U(r_*,\theta)$
What follows is ideas that my lecturer has put forward, which I do not fully understand. I will endeavour to write exactly as my lecturer has so that you may understand.
We can evaluat $U(r,\theta)$ at $r=r_*$ which would lead to $$U(r_*,\theta) = a_0+\sum_{m=1}^{\infty}r_*^m(a_m\cos(m\theta) +b_m\sin(m\theta))=f(\theta)$$ I assume the use of this is so that we can compare the series to the boundary and use that to 'extract' the Fourier coefficients.
Returning to the first equation, it has been given to us that we can find the values of the coefficients as follows: $$a_0 = \frac{1}{2\pi}\int_0^{2\pi}f(\theta)d\theta $$ $$a_m = \frac{1}{\pi r_*^m}\int_0^{2\pi}f(\theta)\cos(m\theta)d\theta $$ $$b_m = \frac{1}{\pi r_*^m}\int_0^{2\pi}f(\theta)\sin(m\theta)d\theta $$
My confusion comes from incorporating the $r_*^m$s in our general solution. In our lecture notes, we have been given two examples and they each treat the inclusion of $r_*$ differently.
Example $1$: Consider the piecewise defined boundary condition $$U(r_*,\theta)=\ \begin{cases} U_1 & 0 < \theta \leq \pi \\ U_2 & \pi <\theta <2\pi \end{cases} \ $$ Where $U_1,U_2 \in \mathbb{R}$. We can use the above integrals to solve for the Fourier coefficients. I will only focus here on $b_m$ as $a_0$ is easy to compute and $a_m=0$.
My lecturers solution has said that $$b_m =\frac{1}{\pi}\int_0^{2\pi}f(\theta)\sin(m\theta)d\theta$$ The details of the integral are not important here, but what I have concerns about is that they have said that $b_m = \frac{1}{\pi}\dots$, shouldn't it be that $b_m = \frac{1}{\pi r_*^m}\dots$, as was specified in the way of deriving the coefficients? This is carried forward and the final solution does not contain $r_*^m$ at all: $$U(r,\theta) = \frac{U_1+U_2}{2} + \sum_{m=1}^{\infty}\frac{r^m}{m}(1-(-1)^m)\sin(m\theta) $$
Example $2$: This time we set $U(r_*,\theta) = \frac{1}{2}+\frac{1}{2}\cos(2\theta)$. This time $a_0 = \frac{1}{2},a_2=-\frac{1}{2r_*^2},a_m=0$ and $b_m=0$. This gives the solution: $$U(r,\theta)=\frac{1}{2}-\frac{r^2}{2r_*^2}\cos(2\theta) $$ Note that this time the $r_*$ is included explicitly.
So that's my problem, is the exclusion of $r_*$ in the first example a mistake, or is there something more complicated that I am missing?