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I'm looking back at some complex analysis and have gotten myself a little muddled in all of the definitions analytic/ regular/ holomorphic/ differentiable/ conformal...

In particular, at the moment I'm thinking about conformal functions $f(z)$ on open sets in $\mathbb{C}$. Many of the conformal maps I'm using are bijections on $\hat{\mathbb{C}}$ e.g. Mobius Transformations. But other conformal maps are only bijections when restricted to a certain set e.g. $f(z)=z^2$ from $\{z:\vert \mathrm{arg}(z) \vert < \frac{\pi}{4} \}$ to $\{z: \vert \mathrm{arg}(z) \vert < \frac{\pi}{2} \}$ is conformal and a bijection but $f(z)=z^2$ from $\{z:\vert \mathrm{arg}(z) \vert < \frac{2\pi}{3} \}$ to $\mathbb{C}-\{0\}$ is not a bijection.

Does bijection-ness feature of any of the definitions of analytic/ regular/ holomorphic/ differentiable/ conformal? Is it a result of any of the definitions? (e.g. might "analytic" imply "bijection"?) Is there a specific name for conformal functions like $f(z)=z^2$ that can be restricted so that they become bijections? Thanks for any help!

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    source A regular function is analytic and single valued. But since analytic function implies taylor series of of this function exists. And since taylor series is single valued. It just means regular = analytic. i.e. no need for single valued condition. – ark Jan 25 '23 at 15:30

2 Answers2

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Firstly, let me clear up the definitions for you. Let $U$ be an open set. Then:

An analytic function on $U$ (real or complex) is infinitely differentiable and equal to its Taylor series in a neighbourhood of every point in $U$.

A holomorphic function on $U$ (necessarily complex) is complex differentiable in a neighbourhood of every point in $U$.

A conformal function is a holomorphic function whose derivative is non-zero on $U$.

Avoid using the term "differentiable function" when working with complex functions. Use one of the terms above instead. I would also avoid "regular": it means the same as "analytic" but isn't well used.

For all functions (real or complex), analytic implies holomorphic. For complex functions, Cauchy proved that holomorphic implies analytic (which I still find astounding)! Hence conformal also implies holomorphic and analytic.

None of the definitions involve a function being a bijection, nor do any of them imply that a function is a bijection. You offer an example yourself: $f(z)=z^2$ from $\{z: \vert \mathrm{arg}(z) \vert < \frac{2\pi}{3} \}$ to $\mathbb{C}-\{0\}$ is conformal, hence also holomorphic and analytic, but is not a bijection.

However, in Latrace's answer to this question they prove that:

If $f$ is conformal in an open set $G$, then for each $a \in G$ there exists an $r > 0$ such that the restriction of $f$ to $D(a;r)$ is one-to-one (where $D(a;r)$ represents the open disk centered at $a$ of radius $r$).

... so in fact every conformal function can be restricted so that they become a bijection.

Malkin
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None of these types of functions have to be bijective. However, for every point $z_0$ of the domain of any conformal map $f$, there is a neighborhood $U$ of $z_0$ such that the restriction of $f$ to $U$ is injective.

On the other hand, note that constant functions are analytic, holomorphic and differentiable.

José Carlos Santos
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