I am trying to prove that holomorphic mappings on $\mathbb{C}$ are angle preserving. Could anyone tell me whether this proof is valid, and, if there are any mistakes provide explanations?
Statement: Holomorphic mappings on $\mathbb{C}$ preserve all angles of intersection.
Proof: Let $f:\mathbb{C}\to\mathbb{C}$ be a holomorphic function on $\mathbb{C}$. This implies that there exists two functions $u,v:\mathbb{R}^2\to\mathbb{R}^2$ of class $\mathcal{C}^1$ such that $\forall(x,y)\in\mathbb{R}$ we have $f(z=x+iy)=u(x)+iv(x)$.
The Jacobian matrix associated with this non-linear transformation is consequently: $$\mathcal{M_J}=\begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x}\\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} \end{pmatrix}$$ $f$ is holomorphic on $\mathbb{C}$, so in other words, if we state that: $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=a$$ $$-\frac{\partial v}{\partial x}=\frac{\partial u}{\partial y}=b$$ We can re-write our matrix as: $$\mathcal{M_J}=\begin{pmatrix} a & -b\\ b & a \end{pmatrix}$$ Where $(a,b)\in\mathbb{R}$ since the partial derivatives are real valued.
Now let us consider $(\vec{i},\vec{j})\in\mathbb{R}^{2\times 2}\times\mathbb{R}^{2\times 2}\backslash \begin{pmatrix} 0\\ 0 \end{pmatrix}$. After having applied the mapping of $f$ onto the plane these two vectors become $\vec{I}$ and $\vec(J)$: $$\vec{I}=\mathcal{M_J}\vec{i}=\begin{pmatrix} a & -b\\ b & a \end{pmatrix}\begin{pmatrix} A\\ B \end{pmatrix}=\begin{pmatrix} aA-bB\\ Ab+aB \end{pmatrix}$$ $$\vec{J}=\mathcal{M_J}\vec{j}=\begin{pmatrix} a & -b\\ b & a \end{pmatrix}\begin{pmatrix} C\\ D \end{pmatrix}=\begin{pmatrix} aC-bD\\ aD+bC \end{pmatrix}$$ $A,B,C$ and $D$ being real valued. To find the original angle $\theta$ between these two vectors before the mapping, we simply have to compute the dot product divided by the multiplication of their norms and apply the $\arccos$ function to the lot. However, because $\arccos$ is a bijection of the interval $[-1;1]\to[0;\pi]$ we only need to verify that: $$\frac{\vec{i}\cdot\vec{j}}{||\vec{i}||\times||\vec{j}||}=\frac{\vec{I}\cdot\vec{J}}{||\vec{I}||\times||\vec{J}||}$$ To conclude that $\theta=\varphi$, where $\varphi$ is the angle between $\vec{I}$ and $\vec{J}$. Hence our problem has become a lot easier to solving. Let us compute the dot product between $\vec{I}$ and $\vec{J}$: $$\vec{I}\cdot\vec{J}=(aA-bB)(aC-bD)+(bA+aB)(bC+aD)$$ $$=a^2AC-abAD-abBC+b^2BD+b^2AC+abAD+abBC+a^2BD$$ $$=AC(a^2+b^2)+BD(a^2+b^2)$$ $$=(a^2+b^2)(AC+BD)$$ $$=(a^2+b^2)(\vec{i}\cdot\vec{j})$$ Now for the multiplication of the norms of the vectors, which we will do squared to simplify the expression, taking the square root later: $$(||\vec{I}||\times||\vec{J}||)^2=((Aa-bB)^2+(bA+aB)^2)((aC-bD)^2+(bC+aD)^2)$$ $$=(A^2a^2-2abAB+b^2B^2+b^2A^2+2abAB+a^2B^2)(a^2C^2-2abCD+b^2D^2+b^2C^2+2abCD+a^2D^2)$$ $$=(A^2(a^2+b^2)+B^2(a^2+b^2))(C^2(a^2+b^2)+D^2(a^2+b^2))$$ $$=(a^2+b^2)^2(A^2+B^2)(C^2+D^2)$$ $$\Longleftrightarrow ||\vec{I}||\times||\vec{J}||=(a^2+b^2)(||\vec{i}||\times||\vec{j}||)$$ Hence taking the fraction yields: $$\frac{\vec{I}\cdot\vec{J}}{||\vec{I}||\times||\vec{J}||}=\frac{(a^2+b^2)\vec{i}\cdot\vec{j}}{(a^2+b^2)||\vec{i}||\times||\vec{j}||}=\frac{\vec{i}\cdot\vec{j}}{||\vec{i}||\times||\vec{j}||}$$ Which concludes our proof as this is what we sought to find.