Intereting Posts

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Divisor in $\mathbb{C}$ $\implies$ divisor in $\mathbb{R}$?

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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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.

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