# Conformal and Real Differentiable function implies entire function

I want to show $f$ is entire given the following conditions: $f:\mathbb{C}\to\mathbb{C}$, $f$ is real differentiable, and $f$ is conformal when $f'$ is not $0$.

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Real differentiability at ${\bf z}_0\in{\mathbb R}^2$ means that one has $$f({\bf z}_0+{\bf Z})-f({\bf z}_0)=L.{\bf Z}+o(|{\bf Z}|)\qquad({\bf Z}\to{\bf 0}),$$ where $L$ is a certain linear map. In the case at hand this $L$ necessarily has the form $$\bigl[L\bigr]=\lambda\left[\matrix{\cos\phi & -\sin\phi \cr \sin\phi& \cos\phi \cr}\right]=:\left[\matrix{a& -b \cr b& a\cr}\right]\ ,\qquad \lambda\geq0, \quad a,b\in{\mathbb R}\ .$$ Changing from vectorial notation to complex notation we therefore have $$f(z_0+Z)-f(z_0)=(aX-bY)+i(bX+aY)+o(|Z|)=(a+ib) Z+o(|Z|)\qquad (Z\to0)\ ,$$ and this implies that $f$ has a complex derivative at $z_0$ given by $f'(z_0)=a+ib$ .
Since this holds for all $z_0\in{\mathbb C}$ the given function $f$ is indeed entire, and we didn't even need to assume that $f$ is $C^1$.