Conformal invariance of harmonic functions

Remarks

Proof idea: near a point $\varphi(z_0)$, harmonic conjugates exist locally, so $h = \operatorname{Re} F$ for a holomorphic $F$ on a disc around $\varphi(z_0)$. Then $h \circ \varphi = \operatorname{Re}(F \circ \varphi)$ near $z_0$, and $F \circ \varphi$ is holomorphic. A direct computation gives $\Delta (h \circ \varphi) = \abs{\varphi'}^2 \, (\Delta h) \circ \varphi$, where $\Delta$ is the Laplacian. Example: $h(w) = \log\abs{w}$ is harmonic on $\C \setminus \set{0}$, and with $\varphi(z) = z^2$ one gets $h(\varphi(z)) = 2 \log\abs{z}$, again harmonic. Holomorphy of $\varphi$ is needed: for the smooth map $\varphi(x, y) = (x, y^2)$ and $h(s, t) = s^2 - t^2$, the function $h \circ \varphi = x^2 - y^4$ has Laplacian $2 - 12 y^2 \ne 0$. The statement also holds for antiholomorphic $\varphi$, because $h \circ \bar\varphi$ equals $\tilde h \circ \varphi$ with $\tilde h(w) = h(\bar w)$, which is harmonic.

Going deeper

For a student

This is the basis of conformal methods in planar potential theory. With the Riemann mapping theorem, the Dirichlet problem on a simply connected domain reduces to the disc, where the Poisson integral solves it. The same invariance makes harmonic measure and Green's functions conformally invariant. It also shows that harmonicity makes sense on a Riemann surface, since the transition maps between charts are holomorphic.