# Complex Mapping - Is transforming boundaries enough?

1. Sep 16, 2015

### davidbenari

Say I will make the transformation from the $z$ plane to the $w$ plane. Moreover, I'll transform a region $R$ with boundary $C$ in the $z$ plane to something in the $w$ plane.

Why is it that if I know the equations for $C$ then I can transform these and immediately know that $R$ will be inside the transformed boundaries? Why isn't it the case that some point in $R$ maps into some point not inside $C'$ (which is my transformed boundary)?

Thanks.

2. Sep 16, 2015

### RUber

You have to say something about the transformation. Normally, you are using linear, or at least continuous mappings which by definition will preserve this inside-outside relationship.

3. Sep 16, 2015

### davidbenari

According to what you say I think continuous mappings would be what I'm dealing with. Strangely I didn't think of continuity, which makes this more intuitive, but not obvious to me.

I see that every point in the neighborhood of $z_o$ in $R$ will also be mapped in $R'$ as a simple closed region. I guess I can visualize these small neighborhoods expanding until they reach the transformed contour...

Is this the argument?

I was told this was because analytic transformations are "open maps" is this what they mean by it?

4. Sep 16, 2015

### Svein

Counterexample: Let $C= (z:\lvert z \lvert = 1)$, then $R= (z:\lvert z \lvert < 1)$. Now transform this using $w=\frac{1}{z}$. This maps C onto C, but R is mapped to the outside of C.

5. Sep 16, 2015

### davidbenari

Svein, but if the function $w$ is analytic in every point in $R$ then what I say will hold right?

6. Sep 16, 2015

### Svein

Sorry, I do not know. I checked out Ahlfors, and he carefully does not make that statement. It depends very much on C and the concept of inside.