= Solution
Assume the image of the domain of $f$ lies in the domain where $g$ is <holomorphic>. The complex <chain rule> gives
$$
\boxed{(g\circ f)'(z)=g'(f(z))f'(z),}
$$
so the composition is holomorphic, including at points where $f'=0$. Its real part $S(x,y)=s(u(x,y),v(x,y))$ is therefore a <harmonic function>.
A direct verification uses the <composition of harmonic conjugate pairs>. The multivariable <chain rule> gives
$$
\Delta S=s_{uu}|\nabla u|^2+2s_{uv}\nabla u\cdot\nabla v+s_{vv}|\nabla v|^2+s_u\Delta u+s_v\Delta v.
$$
The <Cauchy-Riemann equations> imply $|\nabla u|^2=|\nabla v|^2$; parts (i) and (ii) remove the last three terms. Since $s$ is harmonic in its own variables,
$$
\boxed{\Delta S=|\nabla u|^2(s_{uu}+s_{vv})=0.}
$$
Back to article page
