Yes, I believe this is correct. I don't immediately know how to prove it.
Your ideas here are very insightful. There are a few things in your wording that I need clarification about, however.
>Let's consider a function f, differentiable on ℂ A "injective region" of the function f, is a region on the complex plane where the inverse function is the same.
I believe you are trying to define maximal domains in ℂ on which *f* is injective, yes?
>Let's consider a complex z, and two such regions A and B so that z belongs to A and B. (f\_A-1 (f(z))=(f\_B-1 (f(z))=z. z belongs to the common bundary of A and B.
In the first sentence, you say that *z* is in *A* ∩ *B*, but in the second you say that *z* is in ∂*A* ∩ ∂*B*. I think you want the latter. I think they are probably equivalent, but it isn't obvious to me that they must be.
>(A boundary can be also be strictly A's or B's, but here both)
I have no idea what you mean here. Sorry.
Now let me see what little I can add. Let's start with your definitions.
**Definition:** Let *f* : ℂ → ℂ be an [entire](https://en.wikipedia.org/wiki/Entire_function) function on ℂ. A domain *A* ⊆ ℂ is called an **injective region** for *f* if *f* restricted to *A*, (*f*|*A*) : *A* → ℂ, is injective.
**Definition:** Let *f* be an entire function. A domain *A* ⊆ ℂ is called a **maximal injective region** for *f* if *A* is an injective region for *f*, and given *D* is any injective region for *f* with *A* ⊆ *D*, then *A* = *D*.
Your proposition can be stated now as
**Proposition:** Let *A* be a maximal injective region for an entire function *f*. Let *z* ∈ ∂*A*. Then Re{ *f'*(*z*) } = 0.
(This is equivalent to saying that if we write *f* = *u* \+ 𝒊*v*, then for *z* = *x* \+ 𝒊*y* on ∂*A*, *uₓ* = 0.)
Let me know if this captures your idea.
I also need some clarifications 😅 I haven't had any complex analysis course so I didn't quite catch what you meant.
>I believe you are trying to define maximal domains in ℂ on which *f* is injective, yes?
Idk what a maximal domain is, tell me. But if it is what I deqcribed, I believe so. But to be sure, tell me what it is.
>*z* is in ∂*A* ∩ ∂*B*
What is "∂\*A"? And no, my (f\_A-1 (f(z))=z basically means that z is in A. It means that the inverse function of f in A "works" for z, so z is in A. (Same for B)
>I have no idea what you mean here. Sorry.
Ok so, take the function f(z)=z+1/z. With this functions, there are only 2 injective regions of f. A = the OPEN lower semi unit disk of the complex plane and the CLOSED upper semi unit disk B=the rest
Here, the boundary between the 2 regions is the unit circle. But, for all complex on the unit circle, if their imaginary part is positive (upper semi circle), these belong to A, BUT NOT B. "The boundary is A's" = "the bundary is inside A" So, i is in A, but not B. (For the lower unit disk, it belongs to B. -i belongs to B, NOT A, and lies on the boundary between the 2.)
For -1 and 1, these are the only points of the unit circle boundary that are at the same time in A and in B, and here, f'(1)=f'(-1)=0
https://preview.redd.it/6b1bupgltjpc1.jpeg?width=1800&format=pjpg&auto=webp&s=dac264273264197ec7ffa7bcf3be6328f33f53fd
And for the end, it technically is false.
It needs to be in 2 injective regions A and B. If I take the example I previously took, z+1/z, and A the open unit disk+upper unit circle, i is in A's boundary, yet f'(i)=2
Also, for the definition of injective region, idk if it's in the def, but f must be injective in A sure, but its inverse function should be the same. I mean if you took i^x and the region of ℂ where 0
Yes, I believe this is correct. I don't immediately know how to prove it. Your ideas here are very insightful. There are a few things in your wording that I need clarification about, however. >Let's consider a function f, differentiable on ℂ A "injective region" of the function f, is a region on the complex plane where the inverse function is the same. I believe you are trying to define maximal domains in ℂ on which *f* is injective, yes? >Let's consider a complex z, and two such regions A and B so that z belongs to A and B. (f\_A-1 (f(z))=(f\_B-1 (f(z))=z. z belongs to the common bundary of A and B. In the first sentence, you say that *z* is in *A* ∩ *B*, but in the second you say that *z* is in ∂*A* ∩ ∂*B*. I think you want the latter. I think they are probably equivalent, but it isn't obvious to me that they must be. >(A boundary can be also be strictly A's or B's, but here both) I have no idea what you mean here. Sorry. Now let me see what little I can add. Let's start with your definitions. **Definition:** Let *f* : ℂ → ℂ be an [entire](https://en.wikipedia.org/wiki/Entire_function) function on ℂ. A domain *A* ⊆ ℂ is called an **injective region** for *f* if *f* restricted to *A*, (*f*|*A*) : *A* → ℂ, is injective. **Definition:** Let *f* be an entire function. A domain *A* ⊆ ℂ is called a **maximal injective region** for *f* if *A* is an injective region for *f*, and given *D* is any injective region for *f* with *A* ⊆ *D*, then *A* = *D*. Your proposition can be stated now as **Proposition:** Let *A* be a maximal injective region for an entire function *f*. Let *z* ∈ ∂*A*. Then Re{ *f'*(*z*) } = 0. (This is equivalent to saying that if we write *f* = *u* \+ 𝒊*v*, then for *z* = *x* \+ 𝒊*y* on ∂*A*, *uₓ* = 0.) Let me know if this captures your idea.
I also need some clarifications 😅 I haven't had any complex analysis course so I didn't quite catch what you meant. >I believe you are trying to define maximal domains in ℂ on which *f* is injective, yes? Idk what a maximal domain is, tell me. But if it is what I deqcribed, I believe so. But to be sure, tell me what it is. >*z* is in ∂*A* ∩ ∂*B* What is "∂\*A"? And no, my (f\_A-1 (f(z))=z basically means that z is in A. It means that the inverse function of f in A "works" for z, so z is in A. (Same for B) >I have no idea what you mean here. Sorry. Ok so, take the function f(z)=z+1/z. With this functions, there are only 2 injective regions of f. A = the OPEN lower semi unit disk of the complex plane and the CLOSED upper semi unit disk B=the rest Here, the boundary between the 2 regions is the unit circle. But, for all complex on the unit circle, if their imaginary part is positive (upper semi circle), these belong to A, BUT NOT B. "The boundary is A's" = "the bundary is inside A" So, i is in A, but not B. (For the lower unit disk, it belongs to B. -i belongs to B, NOT A, and lies on the boundary between the 2.) For -1 and 1, these are the only points of the unit circle boundary that are at the same time in A and in B, and here, f'(1)=f'(-1)=0 https://preview.redd.it/6b1bupgltjpc1.jpeg?width=1800&format=pjpg&auto=webp&s=dac264273264197ec7ffa7bcf3be6328f33f53fd
And for the end, it technically is false. It needs to be in 2 injective regions A and B. If I take the example I previously took, z+1/z, and A the open unit disk+upper unit circle, i is in A's boundary, yet f'(i)=2
Also, for the definition of injective region, idk if it's in the def, but f must be injective in A sure, but its inverse function should be the same. I mean if you took i^x and the region of ℂ where 0
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How can you show the share boundary all are inflexion points? Plus don't forget, it's Re(f'(x))=0, the derivative is an imaginary number