# 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??

• If $\partial_z f(z)$ is always non-negative, then it is constant, and $f(z) = c\cdot z$. I guess you meant $\partial_z f(z)$ always non-zero. – Daniel Fischer Jan 17 '14 at 14:19
• Yes I meant non-zero, but why should it be constant by just being non-negative? – user119139 Jan 17 '14 at 14:40
• Because a real valued holomorphic function is constant (on a domain). Non-negative implies real. – Daniel Fischer Jan 17 '14 at 14:42
• Oh right, by Cauchy-Riemann – user119139 Jan 17 '14 at 14:44

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.