[Complex Analysis] prove non-existence of conformal map

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Homework Help Overview

The discussion revolves around proving the non-existence of a conformal map from the open unit disk D(0,1) to the complex plane \mathbb{C}. Participants are exploring the properties of conformal maps, particularly in relation to their bijectiveness and the implications of the boundary of D(0,1).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning whether a conformal map must be bijective and discussing the implications of this on the existence of such a map. There are attempts to connect the properties of conformal maps with the concepts of boundedness and differentiability.

Discussion Status

The discussion is ongoing, with participants raising important questions about the definitions and properties of conformal maps. Some guidance has been offered regarding the relationship between conformality, analyticity, and the nature of the mappings involved.

Contextual Notes

There is a mention of differing interpretations of conformal maps in the context of the course material, specifically regarding whether they must be bijective or merely locally invertible. This highlights potential confusion stemming from the course content.

nonequilibrium
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Homework Statement


"Show that there is no conformal map from D(0,1) to \mathbb C"
and D(0,1) means the (open) unit disk

Homework Equations


Conformal maps preserve angles

The Attempt at a Solution


I don't have a clue. I thought the clou might be that D(0,1) has a boundary, and C doesn't, so I tried to draw some circles/lines on D(0,1) that couldn't be imaged onto C whilst preserving angles, but to no avail.

Homework Statement

 
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I'm guessing you mean a bijective function. Then the inverse map would map C to D(0,1). Think about what you might conclude from that.
 
Last edited:
Oh, does a conformal map have to be bijective?

(in that case: the inverse would be bounded ==(Liouville)==> constant, contradiction)
(Now I'm also assuming differentiability! So does a conformal map have to be bijective AND/OR analytical?)
 
mr. vodka said:
Oh, does a conformal map have to be bijective?

(in that case: the inverse would be bounded ==(Liouville)==> constant, contradiction)
(Now I'm also assuming differentiability! So does a conformal map have to be bijective AND/OR analytical?)

You ask good questions. Those are the questions I was asking myself after I posted. Looking things up, indeed a map in C is conformal iff it's analytic and it has a nonzero derivative. I'm still working on justifying my reckless post. Anything in your course to help?
 
the confusing this: in my course itself (the free course by ash & novinger), it says a conformal map only has to be locally invertible (which is actually a consequence of the non-vanishing derivative). But in pen I had written down that a conformal map is a bijection (and the way it was written makes me think it was dictated by the professor)
 

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