Considering two regions and a holomorphic function, show the following 
I'm very lost on how to do this question. Do I use cauchy-reimann equations somehow? Thank you for your help!
 A: We use $\triangle f=4\dfrac{\partial^2f}{\partial z\partial\bar{z}}$.
$$\begin{eqnarray}
\triangle (\psi\circ f) 
&=& 4\dfrac{\partial^2}{\partial z\partial\bar{z}}(\psi\circ f)\\
&=& 4\dfrac{\partial}{\partial z}\left(\dfrac{\partial}{\partial\bar{z}}(\psi\circ f)\right)\\
&=& 4\dfrac{\partial}{\partial z}\left(
\dfrac{\partial}{\partial f}(\psi\circ f)\dfrac{\partial f}{\partial\bar{z}}+\dfrac{\partial}{\partial\bar{f}}(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}}
\right)\\
&=&4\dfrac{\partial}{\partial z}\left(
\dfrac{\partial}{\partial\bar{f}}(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}}
\right)\\
&=&4\dfrac{\partial^2}{\partial z \partial\bar f}
(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}} + 4\dfrac{\partial}{\partial\bar f}
(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}\partial z}
\\
&=&4\dfrac{\partial}{ \partial\bar f}\left( \dfrac{\partial}{ \partial f}(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}} \dfrac{\partial{f}}{\partial{z}} + \dfrac{\partial}{ \partial \bar f}(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}} \dfrac{\partial\bar{f}}{\partial{z}}\right)
 + 4\dfrac{\partial}{\partial\bar f}
(\psi\circ f)\overline{\left(\dfrac{\partial{f}}{\partial\bar{z}\partial z}\right)}
\\
&=&4\dfrac{\partial}{ \partial\bar f}\left( \dfrac{\partial}{ \partial f}(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}} \dfrac{\partial{f}}{\partial{z}}\right)
\\
&=&4\dfrac{\partial}{ \partial\bar f }\left( \dfrac{\partial}{ \partial f}(\psi\circ f)|f'|^2\right)
\\
&=&4\dfrac{\partial}{ \partial\bar f }\left( \dfrac{\partial}{ \partial f}(\psi\circ f)\right)|f'|^2
\\
&=&4\dfrac{\partial^2}{ \partial\bar f \partial f }(\psi\circ f))|f'|^2
\\
&=& \triangle \psi .|f'|^2
\end{eqnarray}
$$
Lines 4 and 7 we use the fact that $\dfrac{\partial f}{ \partial \bar z}=0$ because $f$ holomorphic
