# Domain of bijectivity of function $f:\mathbb{C}\rightarrow\mathbb{C}$

There is a type of problems in my course in Complex analysis that I don't fully understand them.

Given function $f:\mathbb{C}\rightarrow\mathbb{C}$, $f(z)=z^2$. You must specify the analytic and bijective domain of this function.

I'll show you my solution, and then ask the question.

Since $z\in\mathbb{C}$ we can present it in the form of $z=x+iy$, where $x,y\in\mathbb{R}$. Then it's easy to present $f(z)$ it in the form of $f(x,y)=u(x,y)+iv(x,y)$.

$$f(x,y)=(x+iy)^2=x^2-y^2+2iyx$$ So $u(x,y)=x^2-y^2$ and $v(x,y)=2yx$ Then we can check analyticity of function by substituting these functions into Cauchy–Riemann equations .

To check bijective of function $f(z)$ I used this theorem:

Function is bijective if and only if it is invertible.

So I used Inverse function theorem to find point in neighbourhood of which the function is invertible.

For this problem I got that function has the inverse in neighbourhoods of all points except neighbourhood of point $(0,0)$ (Jacobian equal $0$ in this point)

So my question is

• I found properties of function $f(z)$ in the neighborhood of any point, but (as I think) I do not answer for the question of problem. Because knowing the properties in the neighborhood of any point does not give anything about the bijective domain. (You can see it here in "Example" section.) I would be grateful if you would explain to me what to do in such a situation.

Update

I took into account remarks in comments: I need a maximum domain of bijectivity.

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What does "the bijective domain" mean? – Chris Eagle Jan 13 '13 at 17:03
Being invertible isn't sufficient to guarantee that a function is surjective. – Git Gud Jan 13 '13 at 17:08
I mean a set of points on which the function $f(z)$ is bijection. I apologize, because English is not my native language. – Oiale Jan 13 '13 at 17:08
This question is a bit ill-posed. There are lots of domains where $f$ is bijective. – JSchlather Jan 13 '13 at 17:23
There is the problem of articles here. If the question asks for the bijective domain, and that is defined as a domain on which the function is injective, then there are many answers and so the word “the” is misplaced. If you amend the question to ask for any domain on which the function is injective, that makes more sense. It becomes much more interesting, and not much harder in this case, if you ask for a maximal such domain. (Articles sometimes get messed up in translation. And the Russian language doesn't even have them …) – Harald Hanche-Olsen Jan 13 '13 at 17:23

1. Prove that for every $w \neq 0$ there are two points $z \in \mathbb C$ such that $f(z) = w$. We say, that there are two branches above $w$ (also note that $f^{-1}(0) = \{0\}$, indicating that there's something interesting going on around zero).
2. Try to construct an inverse function $f^{-1} : A \subset \mathbb C \setminus \{0\} \to \mathbb C$ from some domain $A \subset \mathbb C$. This amounts, for every $w \in A$, to picking one of the two points from $f^{-1}(w)$. Observe that once you pick such a point, then actually some neighbourhood $U \ni z$ is determined by this choice, otherwise the inverse wouldn't be continuous.
3. By choosing bigger and bigger neighbourhoods, you can continue this process for some time. But you can't enlarge $A$ to whole $\mathbb C$ since this process will eventually lead you to the second branch and tell you to pick a different point in $f^{-1}(w)$ than the one you've picked initially.
4. As an example, set $f^{-1}(1) := 1$ and continue the inverse continuously along the curve $e^{i\phi}$, for $\phi$ from $0$ to $2\pi$. What happens when $\phi \to 2 \pi$?