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??

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