Let $f(z)=z+\frac{\overline{z}}{4}$ where $\left|z\right|<1$. Does the function lift to minimal surface?
$h(z)=z$ , $g(z)=\frac{z}{4}$ since $f(z)=h(z)+\overline{g(z)}$ so $h\text{'}(z)=1$ and $g\text{'}(z)=\frac{1}{4}$ then the second dilatation of $f$ is $\omega =\frac{g\text{'}}{h\text{'}}=\frac{1}{4}={\left(\frac{1}{2}\right)}^{2}$ so $p=1$,$q=\frac{1}{2}$ where $p$ and $q$ Weierstrass-Enneper parameters. ($h\text{'}=p$,$g\text{'}=p{q}^{2}$)
f is lifting to minimal surface since $\omega ={q}^{2}$.
But in class we always found the second dilation respect to $z$ and I'm confused that now is constant function. Is that okay? Any idea will be appreciated.