Help making a proof about conformal mappings rigorous $\newcommand{\Rea}{{\text{Re}}}
\newcommand{\Ima}{{\text{Im}}}$
Here is the question straight of the book: http://bit.ly/JtbjsW

Prove the following: Suppose $f: \Omega \rightarrow \mathbb{C}$ is a complex-valued function, that is real-differentiable at $z_0 \in \Omega$, and $J_f (z_0)\neq 0$. If $f$ preserves angles at $z_0$, then $f$ is holomorphic at $z_0$ with $f′(z_0) \neq 0$.

I know that complex differentiation (holomorphicity) is stronger than real differentiation because in the definition of the complex derivative the limit approaches from all directions, rather than just two as is the case with real differentiation. So it seems here that we want to exploit that fact this function preserves angles to give us this sense of limit from all directions, i.e. by considering arbitrary curves, we have that limit will be the same in all directions in some sense. 
I however have been struggling to make this rigorous. I'm also not sure what the Jacobian being non-zero does (I'm not sure if it is supposed to be nonsingular, but the book has it written as $\neq 0$, so I don't know).
Thanks!
Edit: Here is how I proved the converse, i.e. that holomorphic with non-vanishing first derivative implies angle-preserving. 
Proof of Hint: By definition the dot product $(\cdot,\cdot)$ is defined by $(z,w) = \Rea (z\bar w)$. Thus $$(f'(z_0)\gamma'(t_0),f'(z_0)\eta'(t_0)) =  \Rea(f'(z_0)\gamma'(t_0) \overline{f'(z_0)\eta'(t_0)}) = \Rea(|f'(z_0)|^2\gamma'(t_0)\overline{\eta'(t_0)})$$ and we can just pull out all real numbers which gives $$|f'(z_0)|^2 \Rea(\gamma'(t_0)\overline{\eta'(t_0)}) = |f'(z_0)|^2 (\gamma'(t_0),\eta'(t_0)) $$. Now to prove that $f$ preserves angles at $z_0$ we have to have that for any two smooth curves $\gamma$ and $\eta$ intersecting at $z_0$, the angle formed between the curves $\gamma$ and $\eta$ at $z_0$ equals the angle formed between the curves $f\circ γ$ and $f \circ \eta$ at $f(z_0)$. To define an angle we need both its cosine and sine which are defined by $\frac{(z,w)}{|z||w|}$ and $\frac{(z,-iw)}{|z||w|}$ respectively. So we check that $\frac{((f(\gamma(t_0)))',(f(\eta(t_0)))')}{|(f(\gamma(t_0)))'||(f(\eta(t_0)))'|} = \frac{((\gamma(t_0))',(\eta(t_0))')}{|(\gamma(t_0))'||(\eta(t_0))'|} $ and $\frac{((f(\gamma(t_0)))',-i(f(\eta(t_0)))')}{|(f(\gamma(t_0)))'||(f(\eta(t_0)))'|} = \frac{((\gamma(t_0))',-i(\eta(t_0))')}{|(\gamma(t_0))'||(\eta(t_0))'|}$. Since $f$ is holomorphic and both curves are smooth we can use the chain rule. Also we only care about curves that intersect, so
 $\gamma(t_0)=\eta(t_0)=z_0$. This gives that $$\frac{((f(\gamma(t_0)))',(f(\eta(t_0)))')}{|(f(\gamma(t_0)))'||(f(\eta(t_0)))'|} = \frac{(f'(\gamma(t_0))\gamma'(t_0),f'(\eta(t_0))\eta'(t_0))}{|(f'(\gamma(t_0))\gamma'(t_0)||f'(\eta(t_0))\eta'(t_0)|}$$ Using our fact from the hint and $f'(z_0) \neq 0$ which allows for the simple cancellation:  $$\frac{|f'(z_0)|^2 (\gamma'(t_0),\eta'(t_0))}{|(f'(\gamma(t_0))||\gamma'(t_0)||f'(\eta(t_0))||\eta'(t_0)|} = \frac{((\gamma(t_0))',(\eta(t_0))')}{|(\gamma(t_0))'||(\eta(t_0))'|}. $$ Thus we have the cosine of the angle is preserved. The proof for the sine is identical:
This gives that $$\frac{((f(\gamma(t_0)))', -i (f(\eta(t_0)))')}{|(f(\gamma(t_0)))'||(f(\eta(t_0)))'|} = \frac{(f'(\gamma(t_0))\gamma'(t_0),-i f'(\eta(t_0))\eta'(t_0))}{|(f'(\gamma(t_0))\gamma'(t_0)||f'(\eta(t_0))\eta'(t_0)|}$$ And again: $$\frac{|f'(z_0)|^2 (\gamma'(t_0), -i \eta'(t_0))}{|(f'(\gamma(t_0))||\gamma'(t_0)||f'(\eta(t_0))||\eta'(t_0)|} = \frac{((\gamma(t_0))', -i(\eta(t_0))')}{|(\gamma(t_0))'||(\eta(t_0))'|}. $$ Thus the sine is also preserved and so we have that the angle is preserved. 
 A: The statement as formulated is false. In the first place, a function is holomorphic on an open set and not "at a point $z_0$". Now it is easy to define a real differentiable function $f$ whose $df(z_0)$ at some point $z_0$ is a similarity, but $f$ is nowhere holomorphic. Consider, e.g., the function
$$f(z):= z+\bar z^2= (x+x^2-y^2)+i(y-2 x y)\qquad(z:=x+iy)\ .$$
Here $df(0)={\rm id}$, but $f$ is definitely  not holomorphic.
When the function $f$ fulfills the stated condition in every point $z$ of its domain of definition $\Omega$ then $f$ is indeed holomorphic. This can be shown as follows: Let
$$\bigl[df(z_0)\bigr]=\left[\matrix{a&b \cr c& d \cr}\right]\ , \qquad ad-bc>0\ .$$
If $df(z_0)$ preserves angles then the images of $(1,0)$ and $(0,1)$ must be orthogonal, whence $ab+cd=0$. Now the images of $(1,1)$ and $(-1,1)$ have to be orthogonal, too; and this implies $a^2+c^2=b^2+d^2$. It is now easy to see that the matrix $\bigl[df(z_0)\bigr]$ is of the form
$$\bigl[df(z_0)\bigr]=\left[\matrix{\rho\cos\phi &-\rho\sin\phi \cr \rho\sin\phi& \rho\cos\phi \cr}\right]\ , \qquad \rho>0$$
for some $\phi$, whence is the constant multiple of an orthogonal matrix (of determinant $+1$). It follows that there are numbers $A$ and $B$ such that
$$\bigl[df(z_0)\bigr]=\left[\matrix{A&-B \cr B& A \cr}\right]\ , \qquad A^2+B^2>0\ .$$
Interpreting this in the complex sense this means that we in fact have
$$f(z)-f(z_0)=(A+iB)(z-z_0) + o(z-z_0)\qquad(z\to z_0)\ ,$$
which is the same thing as
$$\lim_{z\to z_0}{f(z)-f(z_0)\over z-z_0}=A+iB\ .$$
