Let $D,G$ be two regions and a function $u\in C^2(G)$ (means that it is twice differentiable with respect to $x,y$ and has continuous partials), a function $\phi$ holomorphic on $D$, with $\phi(D)\subset G$. Prove that $\triangle (u\circ \phi)=\triangle u|\phi'|^2)$, where $\triangle$ is the Laplace operator.
Can anyone indicate how to do those calculations? I was stuck when expanding the left hand side to some expression with $\frac{u}{x},\frac{u}{y}$ terms, and I don't know if this is the right way. Thanks in advance.
I like to write a part of the solution as hint. By $\triangle f=4\dfrac{\partial^2f}{\partial z\partial\bar{z}}$ we see \begin{eqnarray} \triangle (u\circ \phi) &=& 4\dfrac{\partial^2}{\partial z\partial\bar{z}}(u\circ \phi)\\ &=& 4\dfrac{\partial}{\partial z}\left(\dfrac{\partial}{\partial\bar{z}}(u\circ \phi)\right)\\ &=& 4\dfrac{\partial}{\partial z}\left( \dfrac{\partial}{\partial\phi}(u\circ \phi)\dfrac{\partial\phi}{\partial\bar{z}}+\dfrac{\partial}{\partial\bar{\phi}}(u\circ \phi)\dfrac{\partial\bar{\phi}}{\partial\bar{z}} \right)\\ \end{eqnarray} $\phi$ is holomorphic so $\dfrac{\partial\phi}{\partial\bar{z}}=0$ and with notation $\dfrac{\partial\bar{\phi}}{\partial\bar{z}}=\overline{\left(\dfrac{\partial\phi}{\partial z}\right)}$ you can continue the solution.