[Complex Analysis] prove non-existence of conformal map

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SUMMARY

The discussion centers on proving the non-existence of a conformal map from the open unit disk D(0,1) to the complex plane ℂ. Key points include the necessity for conformal maps to be both analytic and have a non-zero derivative, which implies they must be locally invertible. The participants highlight that if such a map were bijective, the inverse would be bounded, leading to a contradiction via Liouville's theorem, which states that bounded entire functions are constant. This contradiction reinforces the conclusion that no conformal map can exist between these two domains.

PREREQUISITES
  • Understanding of conformal maps and their properties
  • Familiarity with the concepts of analytic functions and non-zero derivatives
  • Knowledge of Liouville's theorem in complex analysis
  • Basic principles of topology, particularly regarding open and closed sets
NEXT STEPS
  • Study the properties of analytic functions in complex analysis
  • Explore Liouville's theorem and its implications for bounded functions
  • Research the concept of local invertibility in the context of complex functions
  • Examine the relationship between conformal maps and bijective functions
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Students and educators in complex analysis, mathematicians focusing on conformal mappings, and anyone seeking to deepen their understanding of the relationship between different domains in complex functions.

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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