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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).

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Your definition of being conformal is not quite correct. It should rather be something like: $$ (Df_z \xi, Df_z \eta) = c(z) (\xi,\eta)$$ where $\xi,\eta$ are any vectors in the plane and $c(z)$ some non-negative constant. Writing $f(z)=f(x,y)=(u(x,y),v(x,y))$ we have $$ Df_z = \left(\begin{matrix} u'_x & u'_y\\ v'_x & v'_y\end{matrix} \right)$$ Putting one factor $Df_z$ on the other side by transposing we get the identity $$ (\xi, (Df_z^*) Df_z \eta) = c(z) (\xi,\eta)$$ As this is true for all vectors we must have: $$ (Df_z^*) Df_z = c(z) \left(\begin{matrix} 1 & 0\\ 0 & 1 \end{matrix} \right)$$ Checking out the four equations implied by this identity, we see that the matrix $Df_z$ is either a scalar times a rotation matrix or a scalar times a reflection, i.e. $$ Df_z = \left(\begin{matrix} a & -b\\ b & a \end{matrix} \right) \ \ or \ \ Df_z = \left(\begin{matrix} a & b\\ b & -a \end{matrix} \right) $$ with $a$ and $b$ being any real numbers (and $c=a^2+b^2>0$). 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.

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    How do you prove that the definition of conformal map in the question (that is: angle preserving map) is equivalent to the one in your answer? – 2017-02-02
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    If $U$ is angle preserving then it maps orthogonal vectors to orthogonal vectors. So $(u,v)=0$ iff $(Uu,Uv)=(u,U^*Uv)=0$. Varying the couple $u,v$ you realize that every vector $v$ is an eigenvector of $U^*U$. This implies that $U^*U$ is proportional to the identity. And then as mentioned above you check all possible solutions and find the two above types (rotation or reflection, times constant). – 2017-02-02
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    Why "Varying the couple $u,v$ you realize that every vector $v$ is an eigenvector of $U^\ast U$. This implies that $U^\ast U$ is proportional to the identity."? – 2017-02-04
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    Also, how do we get that if $\frac{(f(u),f(v))}{|f(u)||f(v)|} = \frac{(u,v)}{|u||v|}$, then $Df$ preserves angles too, from which $(Df_z \xi, Df_z \eta) = c(z) (\xi,\eta)$ should follow? – 2017-02-04
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    Take any vector $v$ and let $u$ be perpendicular to $v$. Then $U^*Uv$ is perpendicular to $u$, whence proportional to $v$. Writing $A=U^*U$ we have $A v = \lambda_v v$ for all $v$. Take two independent vectors $v_1,v_2$. Then $A (v_1+v_2)=\lambda_{v_1+v_2} (v_1+v_2) = \lambda_{v_1} v_1+\lambda_{v_2} v_2$ which implies $\lambda_{v_1}=\lambda_{v_2}$. Thus $A=cI$. The identity you mention is not correct and not the definition of a holomorphic map being conformal (as I wrote in the beginning of my answer) – 2017-02-04