Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 9 / 4

For a harmonic function $u$, put $g=u_x-i{u}_y$. Its Wirtinger derivative isso the Cauchy-Riemann equations make $g$ holomorphic. In the simply connected domain it has a single-valued primitiveThe integral is path independent: local holomorphic primitives make its value invariant under path homotopies, and being simply connected makes any two paths with the given endpoints homotopic, as in Question 3.

Since $F^{\prime }=g$, the real part satisfies $(\mathrm{Re}F{)}_x=u_x$ and $(\mathrm{Re}F{)}_y=\mathrm{Re}(ig)=u_y$. Hence $\mathrm{Re}F-u$ is constant on the connected domain. Adding a real constant corrects it, producing a single-valued holomorphic $f$ with $\mathrm{Re}f=u$. The imaginary part is a harmonic conjugate and is unique up to an additive real constant. This is the global harmonic conjugate criterion by vanishing periods in the simply connected case.

Take $u(z)=\log |z|$. 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 $u=\mathrm{Re}f$ for a single-valued holomorphic function, the Cauchy-Riemann equations would give $f^{\prime }=u_x-i{u}_y=1/z$. But for $1<\rho <2$,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. DefineThe series converges locally uniformly outside the unit circle, andThus 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.