Definitions of analytic, regular, holomorphic, differentiable, conformal: what implies what and do any imply that a function is a bijection? 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!
 A: 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.
A: 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.
