Show that an analytic mapping is an open mapping

[Jamin2112]

Homework Statement

As in in title.

Homework Equations

Open mapping: maps open sets to open sets.

The Attempt at a Solution

Not sure.

 

[Dick]

Homework Statement

As in in title.

Homework Equations

Open mapping: maps open sets to open sets.

The Attempt at a Solution

Not sure.

That's a pretty poor attempt. If you don't know how to prove it can you at least give us a few thoughts on why you think it might (or might not) be true?

 

[Jamin2112]

That's a pretty poor attempt. If you don't know how to prove it can you at least give us a few thoughts on why you think it might (or might not) be true?

If a function f(z) is differentiable on a set G, then it is continuous on G. If it continuous, the image of every open set is an open set. Done(?).

 

[Dick]

If a function f(z) is differentiable on a set G, then it is continuous on G. If it continuous, the image of every open set is an open set. Done(?).

Nice that you are trying to think about it. But that's not true in the real numbers. f(x)=x^2 is continuous and differentiable, but it's not open. The proof is going to have to involve special properties of analytic functions.

 

[Jamin2112]

Nice that you are trying to think about it. But that's not true in the real numbers. f(x)=x^2 is continuous and differentiable, but it's not open. The proof is going to have to involve special properties of analytic functions.

Anything to do with "Analytic continuation" that I see on Wikipedia?

 

[Dick]

Anything to do with "Analytic continuation" that I see on Wikipedia?

That's part of it. Rouche's theorem is probably the main ingredient.

 

[Bacle2]

It makes sense to me this way:

An open set in f(z) (say an open ball B(f(z),r), to simplify) is one in which every point in B(f(z),r) is in f(z), i.e., every point in B(z,r) is in the image of f(z). This makes every point in B(f(z),r) a solution to , or a zero of, a certain equation.