Explaining why a composition of a harmonic function and a Mobius transform is harmonic

Question

I'm not sure how to actually show/explain this. I only recall learning harmonic functions being functions that satisfy Laplace's Equation.

Let $f(z)=\frac{1+z}{1-z}$ for all $z\in\mathbb{C}\backslash\{1\}$.
a) Show that $f$ maps $\{z\in\mathbb{C}:|z|<1\}$ onto $\{w\in\mathbb{C}:\mathrm{Re}(w)>0\}$.
b) Explain why $H\circ f$ is harmonic in $\{z\in\mathbb{C}:|z|<1\}$ if $H$ is harmonic in $\{w\in\mathbb{C}:\mathrm{Re}(w)>0\}$.

I've done $(a)$ by simply inverting $f$ and using the condition that $|z|<1$.
However, I'm not sure how this is used (if it is) in part (b).

Answer 1

Whilst $\text{holomorphic}\circ\text{harmonic}$ might not be harmonic (for instance, $\sin\circ\Re$ is not harmonic), it is true that $\text{harmonic}\circ \text{holomorphic}=\text{harmonic}$.

An algebraic way to prove it: call $\partial f=\frac{\partial f}{\partial z},\ \overline\partial f=\frac{\partial f}{\partial\overline z}$ and $\Delta^c=\partial\overline\partial=\overline\partial\partial=\frac14\Delta=\frac14\frac{\partial^2 }{\partial x^2}+\frac14\frac{\partial^2 }{\partial y^2}$. Recall the identities $\overline\partial\left[ \overline f\right]=\overline{\partial f}$ and the chain rule $$\partial[f\circ g]=\partial f\circ g\cdot \partial g+\overline\partial f\circ g\cdot \partial\overline g\\\overline\partial[f\circ g]=\partial f\circ g\cdot \overline\partial g+\overline\partial f\circ g\cdot \overline\partial\overline g$$

\begin{align}\Delta^c[H\circ f]&=\partial\left[\partial H\circ f\cdot\overline \partial f+\overline \partial H\circ f\cdot\overline\partial\left[\overline f\right]\right]=\\&=\partial\overline\partial f\cdot\partial H\circ f+\overline\partial f\cdot\partial\left[\partial H\circ f\right]+\partial\overline\partial\left[\overline f\right]\cdot\overline\partial H\circ f+\overline\partial\left[\overline f\right]\cdot\partial\left[\overline\partial H\circ f\right]=\\(H\text{ harm})&=\Delta^c f\cdot\partial H\circ f+\overline\partial f\cdot\partial f\cdot\partial^2H\circ f+\overline{\Delta^c f}\cdot\overline\partial H\circ f+\overline{\partial f}\cdot\partial\overline f\cdot\overline\partial^2H\circ f=\\&=\Delta^c f\cdot\partial H\circ f+\overline\partial f\cdot\partial f\cdot\partial^2H\circ f+\overline{\Delta^c f}\cdot\overline\partial H\circ f+\overline{\partial f\cdot\overline\partial f}\cdot\overline\partial^2H\circ f\end{align}

And it's apparent that this quantity is $0$ on any $\Omega$ where $f$ is holomorphic (because the condition of holomorphy is $\overline\partial f(z)=0$ for all $z\in \Omega$).