Minimal surfaces under conformal parametrizations - proof verification Hi I am studying minimal surfaces from Kuhnel's Curves - Surfaces Manifolds
The theorem states that: If $f$ is a conformal parametrization, $f$ is a minimal surface if and only if the functions $\phi_{1},\phi_{2}, \phi_{3}$ are holomorphic.
So he defines a surface element $f: U \rightarrow \mathbb{R}^3$ with components $f = (f_{1},f_{2},f_{3})$ and defines the map $\phi(u+iv) = \frac{\partial f}{\partial u}(u,v) - i\frac{\partial f}{\partial v}(u,v)$, which is in components for $j = 1,2,3$
\begin{align*}
\phi_{j} (u +iv) = \frac{\partial f_{j}}{\partial u}(u,v) - i\frac{\partial f_{j}}{\partial v}(u,v)  \ (*)
\end{align*}
For showing the theorem I assume the result: $f$ defines a minimal surface $\iff$ the three component functions $f_{1}, f_{2}, f_{3}$ of $f$ have the relation
\begin{align}
\frac{\partial^2 f_{i}}{\partial u_{1}^2} + \frac{\partial^2 f_{i}}{\partial u_{2}^2} = 0 \ (**)
\end{align}
So all we have to do is take the second derivatives and show that their sum equals zero Now his proof is:
\begin{align*}
\frac{\partial^2 f_{k}}{\partial u^2} &= \frac{\partial}{\partial u}(Re \  \phi_{k}), \ \frac{\partial^2 f_{k}}{\partial v^2} =  - \frac{\partial}{\partial v}(Im \  \phi_{k}), \\
\frac{\partial^2 f_{k}}{\partial u \partial v} &= \frac{\partial}{\partial v}(Re \  \phi_{k}) = - \frac{\partial}{\partial u} (Im \ \phi_{k})
\end{align*}
And from that he concludes that (**) is satisfied
Now firstly I am confused about if he is differentiating (*)? w.r.t. u and v
But in that case I can't really make sense of the calculations, any help would be appreciated 
 A: Starting with the definition: 
$$\phi_k(u,v) = \frac{\partial f}{\partial u}(u,v) - i \frac{\partial f}{\partial v}(u,v)\qquad \mbox{for }k=1,2,3$$
we see that $\phi_k$ is being explicitly written as a sum of real and imaginary parts.  So, 
$$ \mbox{Re}(\phi_k) = \frac{\partial f}{\partial u} $$ and 
$$ \mbox{Im}(\phi_k) = -\frac{\partial f}{\partial v} $$
So if we differentiate both sides with respect to $u$ for the real part and $v$ for the imaginary part, we get
$$ \frac{\partial f}{\partial u}\mbox{Re}(\phi_k) = \frac{\partial^2 f}{\partial u^2} $$ and 
$$ \frac{\partial f}{\partial v}\mbox{Im}(\phi_k) = -\frac{\partial^2 f}{\partial v^2} $$
and for the mixed derivatives (which are equal thanks to us being $C^1$) 
$$ \frac{\partial f}{\partial u}\mbox{Im}(\phi_k) = -\frac{\partial^2 f}{\partial v \partial u} = -\frac{\partial^2 f}{\partial u \partial v} = -\frac{\partial f}{\partial v}\mbox{Re}(\phi_k) $$
Now write out the second derivative and use these relations to substitute in -- you'll find that everything cancels and leaves you with $0$.  All without ever knowing what $f$ and $\phi$ explicitly look like!
