Conformal mapping in a region imply analiticity

Question

I've been reading Needham's book Visual Complex Analysis and stopped at the following theorem: a conformal mapping in a region implies analyticity. Could someone explain this theorem to me better geometrically? Furthermore, another consequence that I didn't understand is the fact that the function and its conjugate cannot be analytic at the same time.

Answer 1

I'll leave out some of the technical details here

Geometrically speaking, conformal maps are locally rotation-dilations. Conformal maps are supposed to preserve angles. To specify the angle of intersection of two intersecting lines, we are really looking at the angle of intersection of their tangents. The tangents of the curves at their point of intersection, and the tangents of the conformally transformed curves, should intersect at the same angle. In order to guarantee that the transformed curves have a tangent in the first place, the conformal map should be (real) differentiable. Now, the tangents of the transformed curve only depend on the constant and linear part of the conformal map. And then the angle only depends on the linear part, so on the differential of the map. The differential is a linear map, and the only way for a linear map to preserve angles is for it to be a rotation-dilation, meaning that it rotates and dilates, nothing else. The matrix representation of such a linear map is always of the form

$$r\begin{pmatrix}\cos\varphi&-\sin\varphi\\ \sin\varphi&\cos\varphi\end{pmatrix}=\begin{pmatrix}a&-b\\b&a\end{pmatrix},$$

Where $r>0$ or equivalently $a^2+b^2\neq0$. Such a map dilates by a factor of $r$ and rotates by the angle $\varphi$. But the second form of the matrix let's us see a different aspect: applying this map to a complex number thought of as a vector is the same as multiplying the number by $a+\mathrm ib$. So the conformal map's linear approximation can be expressed as multiplication by a complex number. But that's the same as being complex differentiable! So conformal maps are complex differentiable. And since this is true at every point, they are in fact holomorphic/analytic. (Another way to see this is that the matrix above contains the partial derivatives of the function, since it is the Jacobian, and then it's easy to verify that the Cauchy-Riemann equations are satisfied).

The corollary about the conjugate goes as follows: let $f$ be analytic and not constant on a connected domain. Then it is conformal on its domain except at the isolated points where it's derivative is $0$. So its conjugate is not conformal anywhere. But if the conjugate were analytic, then it would in fact be conformal except on a set of isolated points where its derivative vanishes (since it can't have zero derivative except at isolated points, otherwise it would be constant, and so $f$ would be constant, which it is not). So the conjugate can't be analytic.

The exception of course happens when $f$ is constant. Then its conjugate is also constant and thus analytic.