It seems like if $U$ is an open subset of the complex plane, $\mathbb{C}$, then a function $$f: U \rightarrow \mathbb{C}$$ is conformal if and only if it is holomorphic and its derivative is everywhere non-zero on $U$.

Can anyone explain why conformal map means holomorphic and vice versa?

Thanks in advance


It depends on how one defines conformal. Traditionally, a map is said to be conformal if the derivative is a conformal transformation (equivalent to the map being angle-preserving).

Under this definition, the conjugate function [$f(x + iy) = x - iy$] is conformal. It is easy to see that it is not holomorphic as it does not respect the Cauchy-Reimann equations (CR).

If angles are preserved with orientation in a conformal map (this is not how it is usually defined), then the claim holds.

A function is holomorphic if and only if it is orientation preserving conformal map

The proof is quite easy. Look at the Jacobian. By using CR, you will be able to show that it is a constant multiplied some matrix of rotation. By the same CR you can also show that matrix has positive determinant, and hence, the map preserves orientation.

The converse follows from the fact that orientation preserving conformal maps respect CR. The matrix of the map must be a constant multiplied by some matrix of rotation that has a positive determinant. When this matrix is seen as the Jacobian of some function, you shall be able to show that the function respects CR.

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  • $\begingroup$ I have safely assumed that the function is differentiable (as a map from $\mathbb{R}^2$ to $\mathbb{R}^2$). I believe that conformal maps should be differentiable, but it is not obvious to me why. Perhaps there is some simple counter-example to it $\endgroup$ – Aalok Thakkar Feb 24 '17 at 3:25

First, conformal at $z_0$ means differentiable at $z_0$, with non zero Jacobian and angles-preserving. With Cauchy-riemann formula, you show that $f'(z_0)\neq0$ implies $J(f)_{z_0} \neq 0$. Then, you take $\gamma_1, \gamma_2$ two curves with origin at $z_0$, and compute the angle $\hat{(f(\gamma_1),f(\gamma_2))}$, by the formula $\hat{(\tau_1,\tau_2)}=Arg \frac{\tau_1'(0)}{\tau_2'(0)}$, and find that f is conformal.

For the converse, let $f$ be a conformal map. We need to prove $d_{z_0}f$ to be $\mathbb{C}$-linear.

$\textbf{Lemma}$: it's enough to prove that $d_{z_0}f$ is a $\mathbb{R}$-linear, non constant, conformal map.

The two first facts are obvious, the third comes from the fact that $f$ is conformal, as well.

For this lemma, you take a $\mathbb{R}$-linear, non constant, conformal map $L$, normalize such that $L(1)=1$, but L is conformal gives $L(i) \in i\mathbb{R}$, and then $L(i)=i$, so $L=id$ up to homothetic transformation, then $L$ is $\mathbb{C}$-linear, and you are done.

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  • 2
    $\begingroup$ antiholomorphic maps are also conformal. $\endgroup$ – Steven Gubkin Feb 25 '14 at 3:58
  • $\begingroup$ Noo, because by angles preserving we mean "oriented angles"! $\endgroup$ – Léo Feb 25 '14 at 9:31

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