A holomorphic function is conformal

I am trying to show that if a function $f = u+iv$ is holomorphic with $\partial_z f(z)$ always non zero, then $f$ is a conformal mapping, i.e. it preserves angles between smooth curves.

If $f$ is holomorphic, by Cauchy-Riemann
$$\begin{vmatrix} u_x & u_y\\ v_x & v_y \end{vmatrix} = \begin{vmatrix} u_x & -v_x\\ v_x & u_x \end{vmatrix} = u_x^2 + v_x^2 = |\partial_z f|^2 \neq 0,$$
so changing variables $r, \theta$ s.t.
$$r = |\partial _z f(z)|, \cos \theta = \dfrac{u_x}{|\partial_z f|}, \sin \theta = \dfrac{v_x}{|\partial_z f|}$$
the jacobian matrix of $f$ becomes
$$\begin{pmatrix} u_x & u_y\\ v_x & v_y \end{pmatrix} = r \begin{pmatrix} \cos \theta & – \sin \theta\\ \sin \theta & \cos \theta \end{pmatrix}.$$
Now the Jacobian indeed preserves angles since it is a composition of a rotation with a dilation. But why $f$ should also preserve angles??

Solutions Collecting From Web of "A holomorphic function is conformal"

From $f(z)-f(z_0)=(z-z_0)f'(z_0) + o(|z-z_0|)$ we see that $f$ “hardly differs” from a multiplication with the nonzero complex number $f'(z_0)$. The little-o is really small (by definition) and lets the distinction between curves through $z_0$ of $f(z_0)$ and their tangents vanish.