= Solution
Use <separation of variables> in <Laplace's equation>: writing $\phi=R(r)\Theta(\theta)$ gives
$$
\frac{r^2R''+rR'}R=-\frac{\Theta''}{\Theta}=n^2.
$$
Single-valuedness requires $2\pi$-periodicity, so the nonconstant angular <Fourier modes> have integer $n\ge1$ and are $\cos n\theta,\sin n\theta$. Their radial solutions are $r^n,r^{-n}$. The zero angular mode has radial solutions $1,\log r$. These give the <harmonic Fourier expansions in planar concentric domains> used below.
For the <annulus> with $0<a<b$, constant boundary data suggest a radial <harmonic function> $A+B\log r$. Enforcing both boundary values gives
$$
\boxed{\phi(r,\theta)=1+\frac{\log(r/a)}{\log(b/a)}.}
$$
This has the required boundary values and satisfies $(r\phi_r)_r=0$. It is unique: the difference of two continuous solutions is <harmonic> in the <annulus> and zero on both boundary circles, so the <maximum principle for harmonic functions>, applied to the difference and its negative, makes it zero.
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