= Solution
Take \b[$u(z)=\log|z|$]. In <polar coordinates> its <Laplacian> is $u_{rr}+r^{-1}u_r=-r^{-2}+r^{-2}=0$, so it is <harmonic> on the <annulus>. If $u=\operatorname{Re}f$ for a single-valued <holomorphic function>, the <Cauchy-Riemann equations> would give $f'=u_x-iu_y=1/z$. But for $1<\rho<2$,
$$
\oint_{|z|=\rho}f'(z)\,dz=0,\qquad
\oint_{|z|=\rho}\frac{dz}{z}=2\pi i.
$$
The first equality follows by evaluating the primitive $f$ around a closed curve. This contradiction verifies the absence of a global <harmonic conjugate>.
For the final function, the quotient of squared distances is $|z+1|^2/|z-1|^2$. On $|z|>1$, both $1+1/z$ and $1-1/z$ lie in the open right half-plane, so their <principal logarithms> are single-valued and holomorphic. Define
$$
\boxed{f(z)=2\operatorname{Log}(1+1/z)-2\operatorname{Log}(1-1/z)
=4\sum_{m=0}^{\infty}\frac{z^{-(2m+1)}}{2m+1}.}
$$
The series converges locally uniformly outside the unit <circle>, and
$$
\operatorname{Re}f(z)=2\log|z+1|-2\log|z-1|
=\log\frac{(x+1)^2+y^2}{(x-1)^2+y^2}.
$$
Thus this is the required single-valued <holomorphic function>, even on the larger domain $|z|>1$. Its <derivative> is $2/(z+1)-2/(z-1)$, whose two periods around the holes cancel. The <logarithmic cancellation on an exterior annulus> explains why this example succeeds while $\log|z|$ fails; using independent multivalued logarithms of $z+1$ and $z-1$ would conceal that cancellation.
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