Let $f(x,y)=u(x,y)+iv(x,y)$ be defined on
$$\Omega \rightarrow D$$
where $\Omega,D$ are domains. Let $H(\zeta,\eta)$ be a real valued function defined on $D$ where $w=\zeta + i \eta$ Prove the composite function defined via
$$h(z)=H \circ f(z)$$
defined on $\Omega$ satisfies
$$\frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial y^2}=\vert f'(z) \vert^2 \Delta H$$
where $\Delta H$ is the $2$ dimensional laplacian of $H$. First of all I don't understand why $H$ is real valued if it maps $(\zeta,\eta)$ to $\zeta + i \eta$. Secondly do I use the fact that $f'(z)=u_x+iv_x$? A little stuck on this one. I know I need to use the chain rule somewhere, am I right? Is the following the right approach
\begin{align} \frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial y^2} &= \frac{\partial^2 (H(f(z))}{\partial x^2}+\frac{\partial^2 (H(f(z))}{\partial y^2}\\\\ &=\frac{\partial}{\partial x}(\frac{\partial}{\partial x}(H(f(z)))+\frac{\partial}{\partial y}(\frac{\partial}{\partial y}(H(f(z))) \end{align}
then using chain rule? Can anybody help me on this please?!
Suppose that $f(z)$ is analytic. Otherwise $f'(z)$ has no meaning. Let $w=f(z)$ and then $$ \frac{\partial w}{\partial \bar z}=\frac{\partial \bar w}{\partial z}=0, \frac{\partial^2 w}{\partial z\partial \bar z}= \frac{\partial^2 \bar w}{\partial z\partial \bar z}=0 $$ Noting $$ \Delta f(z)=4\partial_z\partial_{\bar z}f(z) $$ one has \begin{eqnarray} \Delta h&=&4\partial_z\partial_{\bar z}H(f((z))\\ &=&4\partial_z\partial_{\bar z}H(w)\\ &=&4\partial_z\bigg(\frac{\partial H(w)}{\partial w}\cdot\frac{\partial w}{\partial \bar z}+\frac{\partial H(w)}{\partial \bar w}\cdot\frac{\partial \bar w}{\partial \bar z}\bigg)\\ &=&4\partial_z\bigg(\frac{\partial H(w)}{\partial \bar w}\cdot\frac{\partial \bar w}{\partial \bar z}\bigg)\\ &=&4\bigg( \frac{\partial^2 H(w)}{\partial w\partial \bar w}\cdot\frac{\partial w}{\partial z}\cdot\frac{\partial \bar w}{\partial \bar z}+\frac{\partial^2 H(w)}{\partial \bar w^2}\cdot\frac{\partial\bar w}{\partial z}\cdot\frac{\partial \bar w}{\partial \bar z}+\frac{\partial H(w)}{\partial \bar w}\cdot\frac{\partial^2 \bar w}{\partial z\partial \bar z}\bigg) \\ &=&4\frac{\partial^2 H(w)}{\partial w\partial \bar w}\cdot\frac{\partial w}{\partial z}\cdot\frac{\partial \bar w}{\partial \bar z} \\ &=&\vert f'(z) \vert^2 \Delta H. \end{eqnarray}