Question

Need help in complex analysis..

Answer 1

Question in attachment

Answer 2

@ganeshie8

Answer 3

@ljetibo

Answer 4

@Kainui

Answer 5

@welshfella

Answer 6

hmmm. sorry I can't recall this stuff well enough to answer you.

Answer 7

Its been a long long time....

Answer 8

@mathmale

Answer 9

@mathmate

Answer 10

@IrishBoy123

Answer 11

So you have a holomorphic function f on the entire C. B is some subset of that C therefore f is holomophic on it as well.

If you have a holomorfic function then it will necessarily be an open map. That is to say open sets will be mapped to other open sets.

Therefore interior points of B will be interior points of f(B), and the only thing the boundary points of the image of B can be is the image of boundary points of B.

In other words, the boundary of the image of B will be contained in the image of boundary of B, in yet another words the boundary will be mapped to a boundary, except that this mapping isn't necessarily bijective so there's a lot of z element B that will map to some z1 element on the boundary of the image of B but all points w element boundary of B will be mapped to some w1 element boundary of image of B.

I'd answer i) because I have no idea how bounded or unbounded sets would change this. Also please don't mass tag, you've barely asked the question 30 minutes ago. Also I do not guarantee that any of this is correct, it's been a long time since I did any complex analysis and even then it was just the basic basic things that we needed as physicists to expand functions and calculate residues to remove singularities.

Answer 12

In complex analysis, the open mapping theorem states that if U is a domain of the complex plane C and f : U → C is a non-constant holomorphic function, then f is an open map (i.e. it sends open subsets of U to open subsets of C, and we have invariance of domain.).

Answer 13

@ljetibo

Answer 14

Exactly, therefore the "If you have a holomorfic function then it will necessarily be an open map. That is to say open sets will be mapped to other open sets." stands.

I'm not sure if this is confirmation or a question to give additional clarifications? Once you have that the rest follows logically and boundary of f(B) is the subset of f(boundary B). The question to really answer here is whether bounded or unbounded sets behave differently.

Answer 15

I don't think there would be any difference therefore I vote i); but as I said, it's been a while and I've only ever done the basics.

Answer 16

Thanks. :) @ljetibo