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This is not a homework question. Thank you for your help

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- Thread starter esisk
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- #1

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This is not a homework question. Thank you for your help

- #2

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This is an interesting question.

The Riemannian mapping thm. actually gives a partial answer to that question - it is possible to find such a mapping, if the domain is not all of C, so the statement is seriously doubted - otherwise Riemann would have stated his thm in a more general way

However, I'll try to sketch a counter proof:

Assume that f: C -> D is holomorphic. It is evident that f is then an entire function. What is more, for all z in C it is true that |f(z)|<=1, i.e. f is bounded, because D is bounded. Now the Liouville's thm implies that f must be constant, implying the statement is incorrect.

(the Liouville's thm is that very nice tool also used for proving the Fundamental Theorem of Algebra in less that 5 lines)

regards,

marin

- #3

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I thank you for the response, but I believe you have misread the question. I am looking for a function from D to C. Not the other way around; and, yes, otherwise Liousville's Theorem would make it impossible.

Again: Can you map the unit disk onto C?

Thanks

- #4

quasar987

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DontoC.

Of course he means the open disk. The disk and the plane are homeomorphic. But not conformally equivalent (as noted, by Liouville's theorem). So what about mapping the open disk onto the complex plane in a many-to-one manner?

- #6

quasar987

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For some reason I didn't even consider that the OP might mean the open disk!

- #7

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And, yes, following the suggestion from Edgar

I can use (z-i)^2 to map the upper falf plane onto C and

Cayley map to map the open disk onto the upper half plane. I believe this will do it. Thsnk you again.

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