I need to solve Laplace equation in a circular ring of radius $a\le r \le b $ with conditions $$u(a,\theta)=f(\theta),\space \space \space u(b,\theta)=0,$$ and, of course, $$ u(r,-\pi)=u(r,\pi), \space \space \space \frac{du}{d\theta}(r,-\pi)=\frac{du}{d\theta}(r,\pi)$$
So with the variable separation system i made $$u(r,\theta)=R(r)\Theta(\theta)$$ Replacing $$\nabla^2u=R''(r)\Theta(\theta)+\frac{1}{r}R'(r)\Theta(\theta)+\frac{1}{r^2}R(r)\Theta''(\theta)=0$$ Multipliyng by $\frac{r^2}{R(r)\Theta(\theta)}$ $$\frac{r^2R''(r)+rR'(r)}{R(r)}=-\frac{\Theta''(\theta)}{\Theta(\theta)}=\lambda$$ Now i have the equations $$(1) \space \space \space \Theta''(\theta)=-\lambda\Theta(\theta)$$ $$(2) \space \space \space r^2R''(r)+rR'(r)=\lambda R(r)$$ I already know that to satisfy (1) $$ \lambda = 0,\space n^2, \space \space \space where \space \space \space n=1,2,3,...$$ $$\Theta(\theta)=C, \space \space \space \Theta(\theta)=\sin n\theta \space \space \space and \space \space \space \Theta(\theta)=\cos n\theta$$ Consequently, replacing the eigenvalues in (2) i arrived to $$R(r)=C_2+C_1 \ln r, \space \space \space R(r)=C_3r^n+C_4r^{-n} \space \space \space n=1,2,3,...$$ Now.. if I apply the boundary condition $u(b,\theta)=0$ $$R(b)=C_2+C_1 \ln b=0 $$ $$R(b)=C_3b^n+C_4b^{-n}=0$$ So... what now? Does this means that $C_2=C_1=C_3=0$? So $R(r)=C_4r^{-n}$?
Recall that $b$ is just some constant whose value is set prior to any of your work, not a variable.
Let us suppose, just for a moment, that $b = 1$. Then your $R(b)$ family is \begin{align*} C_1 + C_2 \cdot 0 &= 0 \\ C_3 \cdot 1 + C_4 \cdot 1 &= 0 \text{,} \end{align*} which just says $C_1 = 0$, $C_2$ is unconstrained, and $C_3 = -C_4$.
Returning to general $b$, you can solve for two constants in terms of the other two and then replace the two solved constants in your $R(r)$ family, which is (up to typos) how J_P got the solution in his comment.