Proof that conformal maps (statisfying certain conditions) are holomorphic/antiholomorphic How can I use the Cauchy-Riemann equations to prove the following result?

Let $\Omega$ be a connected open subset of $\mathbb{C}$. Let $f:\Omega \to \mathbb{C}$ be $C^1$ (in the real sense, as a map from a subset of $\mathbb{R}^2$ to $\mathbb{R}^2$) and assume that the determinant of its Jacobian that is never $0$.
If $f$ is conformal (that is $\frac{(f(u),f(v))}{|f(u)||f(v)|} = \frac{(u,v)}{|u||v|}$, where $(\cdot, \cdot)$ is the inner product) and the Jacobian is always positive then $f$ is holomorphic.
If $f$ is conformal and the Jacobian is always negative then $f$ is antiholomorphic (that is: $\overline{f}$ is holomorphic).

 A: Your definition of being conformal should be for the derivative of the map and not the map itself.
Let $J=Df(z)=\pmatrix{a & c\\ b & d} \in {\rm GL}_2({\Bbb R})$. Conformality implies in particular that orthogonal vectors map to orthogonal vectors. Let $(e_1,e_2)$ be the canonical base in ${\Bbb R}^2$. Then
$ (Je_1,Je_2)=0 $ and $ 0 = (J(e_1+e_2),J(e_1-e_2)) =|Je_1|^2 - |Je_2|^2$
implies respectively that
$$ ac + bd=0  \ \ {\rm and} \ \
 a^2+b^2=c^2+d^2.$$
The only possibilities are $(c,d)=(-b,a)$ or $(c,d)=(b,-a)$.
The matrix $Df_z$ is thus either a scalar times a rotation matrix or a scalar times a reflection, i.e.
$$  Df_z =  \left(\begin{matrix} a & -b\\ b & a \end{matrix} \right)
\ \ or \ \  Df_z =  \left(\begin{matrix} a & b\\ b & -a \end{matrix} \right)
$$ Comparing with the writing of $Df_z$ using derivatives we obtain in the first case the Cauchy Riemann equations and positive $\det Df_z$ (holomorphic case) and in the second case the Cauchy Riemann equations for $u(x,-y)$ and $-v(x,-y)$ and negative   $\det Df_z$ (antiholomorphic case). Working upwards in the above argument we see that the converse is also true: holomorphic or anti-holomorphic implies conformal.
The first argument carries over to any dimension and shows that a conformal matrix is a scalar multiple of an orthogonal matrix.
