- Let h be a harmonic function on the domain
\[ U : = \left\{ z \in \mathbb { C } : | z | > 1 \right\} . \]
Show that there exists a constant \(c \in \mathbb { R }\) and a holomorphic function \(f\) on \(U\) such that \(\mathrm { R e } f ( z ) = h ( z ) + c \log | z |\) for all \(z \in U\)
Solution
1. The holomorphic gradient of \(h\) has a real residue on the exterior annulus.
Write \(z=x+iy\) and put \(q(z)=h_x(z)-ih_y(z)\). The Cauchy–Riemann equations for \(q\) are precisely \(h_{xx}=-h_{yy}\) and \(h_{xy}=h_{yx}\), so harmonicity makes \(q\) holomorphic on \(U\). Its Laurent expansion is \[ q(z)=\sum_{n\in\mathbb Z}a_nz^n\qquad(1<|z|<\infty) \] [@SS03]. For any \(r>1\), its residue coefficient is \[ a_{-1}=\frac1{2\pi i}\int_{|z|=r}q(z)\,dz. \] Along a curve one has \[ q(z)\,dz=(h_x\,dx+h_y\,dy) +i(h_x\,dy-h_y\,dx). \] The real part is \(dh\), whose integral over the closed circle is zero. Thus the integral is purely imaginary, and \(a_{-1}\) is real. Denote this real number by \(a\).
2. Subtracting \(a/z\) gives a holomorphic gradient with a single-valued primitive.
Set \(c=-a\). The function \(q+c/z=q-a/z\) has no \(z^{-1}\) term. Define \[ F(z)=\sum_{\substack{n\in\mathbb Z\\n\ne-1}} \frac{a_n}{n+1}z^{n+1}. \] To justify this definition, fix \(1<r\leq|z|\leq R<\infty\) and choose \(1<s<r\leq R<T\). If \(M_t=\max_{|z|=t}|q(z)|\), the Laurent coefficient formula gives \(|a_n|\leq M_T T^{-n}\) for \(n\geq0\), and \(|a_{-k}|\leq M_s s^k\) for \(k\geq1\) [@SS03]. The positive-power part of the proposed primitive is therefore bounded termwise by \(M_T R(R/T)^n\). For \(k\geq2\), its negative-power terms have modulus at most \(M_s r(s/r)^k\). Both are summable geometric bounds, uniform on the chosen compact annulus. The same coefficient bounds give uniform convergence of the differentiated series there. Termwise differentiation is consequently valid, and \(F\) is holomorphic with \[ F'(z)=q(z)-a/z. \] The integer powers make \(F\) single-valued on all of \(U\); no logarithm branch is assumed.
3. An additive real constant makes the real part equal to the required function.
Put \(H(z)=h(z)+c\log|z|\). On \(U\), \[ (\log|z|)_x-i(\log|z|)_y =\frac{x-iy}{x^2+y^2}=\frac1z. \] Hence \(H_x-iH_y=q+c/z=F'\). For holomorphic \(F\), this means that \(\operatorname{Re}F\) and \(H\) have the same real gradient. Their difference is locally constant, and \(U\) is connected, so it equals one real constant \(d\). Then \(f=F-d\) is holomorphic on \(U\) and satisfies \[ \operatorname{Re}f=h+c\log|z|, \qquad c=-\frac1{2\pi i}\int_{|z|=r}(h_x-ih_y)\,dz\in\mathbb R. \] This is the asserted representation.