[Complex Analysis] prove non-existence of conformal map

[nonequilibrium]

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

 

[Dick]

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.

 

[nonequilibrium]

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?)

 

[Dick]

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?

 

[nonequilibrium]

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)