- #1

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## Homework Statement

As in in title.

## Homework Equations

Open mapping: maps open sets to open sets.

## The Attempt at a Solution

Not sure.

- Thread starter Jamin2112
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- #1

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As in in title.

Open mapping: maps open sets to open sets.

Not sure.

- #2

Dick

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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?## Homework Statement

As in in title.

## Homework Equations

Open mapping: maps open sets to open sets.

## The Attempt at a Solution

Not sure.

- #3

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If a functionThat'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?

- #4

Dick

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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.If a functionf(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(?).

- #5

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

- #6

Dick

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That's part of it. Rouche's theorem is probably the main ingredient.Anything to do with "Analytic continuation" that I see on Wikipedia?

- #7

Bacle2

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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.

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