TITLE: Gauss map is conformal for minimal surfaces.
QUESTION [1 upvotes]: My problem is:
Prove that the Gauss map of a minimal surface $S$ in the Euclidean space $\mathbb{R}^3$ is conformal.
My intuition tells me that this is true but I don't know how to attack the problem. I've looked at the differential of Gauss map but no ideea, also i don't know how to use the fact that $S$ is minimal. 
EDIT: By This  question I know that the gauss map is anti-holomorphic. Thus my question becomes, is any anti-holomorphic function a conformal function? I know that this is true for holomorphic functions.

REPLY [1 votes]: When $N: S\rightarrow \mathbb{S}^2$ is unit out normal, then assume
that $S$ is a minimal surface whose principal curvatures are $k,\
-k$ at a point $p$. Hence $dN : T_pS\rightarrow
T_{N(p)}\mathbb{S}^2$ is a linear map, i.e. $dN $ is a diagonal
matrix $(-k,k)$, which is a composition of reflection and scaling.
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