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We note that there cannot exist bounded one-one holomorphic map $f:\mathbb C \setminus \{0\} \to \mathbb C.$

Put $D=\{z\in \mathbb C: |z|\leq1\}$ (closed disk).

My Question: How to show there exists $f:\mathbb C \setminus D \to \mathbb C$ which is bounded one-one holomorphic?

(I am guessing Riemann mapping theorem may be useful, but I do not know how (here my domain is not simply connected and RMT is true for simply connected domain), and I am unable to think any other theorem of complex analysis which guarantees one-one and boundedness)

Note see the related question here

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    $\begingroup$ What about $z \mapsto \frac{1}{z}$ $\endgroup$ – user4422 Jul 23 '15 at 6:35
  • $\begingroup$ @thanks; I got it; $\endgroup$ – Inquisitive Jul 23 '15 at 6:40
  • $\begingroup$ @user4422: but what will happen if replace $D$ by closed connected and has more than one element, $\endgroup$ – Inquisitive Jul 23 '15 at 6:41
  • $\begingroup$ If it has non-empty interior you can use the same argument. $\endgroup$ – user4422 Jul 23 '15 at 6:43
  • $\begingroup$ @user4422; sorry, I could not follow you; would you explain me bit more basically I am asking this ; thanks $\endgroup$ – Inquisitive Jul 23 '15 at 6:47
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Hint: No need for fancy theorems when there is a very simple explicit example.

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  • $\begingroup$ thanks; Oh, I think, $f(z)=\frac{1}{z}$ will work. But what will happen if we replace $D$ by closed connected which has more than one element. thanks $\endgroup$ – Inquisitive Jul 23 '15 at 6:38

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