A holomorphic function is conformal

Question

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

score 4 · Answer 1 · answered Mar 13 '18 at 17:58

The angle between two curves through a point is measured by means of the tangent vectors to the curves at the point. Then, a map preserves angles (i.e., it is conformal) if the differential of the map preserves angles. The differential of the map is the linear map defined by the Jacobian matrix. As you have said, the Jacobian corresponds to a conformal linear map, thus proving that holomorphic functions are conformal.

In the case of manifolds, a map f between two manifolds M and M’ induces a linear map between the tangent space of M at p and the tangent space of M’ at f(p), which is called the differential of f at p. The map is conformal if the differential is conformal at every point. In the present case both manifolds M and M’ are the set of complex numbers.

score 2 · Accepted Answer · answered Jan 17 '14 at 14:25

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.

Robert McLean MD PhD · Answer 3 · 2020-03-26T00:21:02.610

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Your calculation of the Jacobian shows the tangent vectors $e_1=\frac{\partial}{\partial x}$ and $e_2=\frac{\partial}{\partial y}$ are rotated and scaled by a common factor. The group of such transformations are called $\textrm{CO}(2)$.These are precisely the angle preserving transformations. Why? Because of the formula

$$\cos(\theta)=\frac{U\cdot V}{\|U\|\|V\|}$$

The scalar term will factor out.

edited Mar 26 '20 at 00:21

answered Mar 23 '20 at 03:27

Robert McLean MD PhD

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A holomorphic function is conformal

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