$\exists f:\mathbb C \setminus D \to \mathbb C$ is bounded one-one holomorphic, how?

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

• What about $z \mapsto \frac{1}{z}$ – user4422 Jul 23 '15 at 6:35
• @thanks; I got it; – Inquisitive Jul 23 '15 at 6:40
• @user4422: but what will happen if replace $D$ by closed connected and has more than one element, – Inquisitive Jul 23 '15 at 6:41
• If it has non-empty interior you can use the same argument. – user4422 Jul 23 '15 at 6:43
• @user4422; sorry, I could not follow you; would you explain me bit more basically I am asking this ; thanks – Inquisitive Jul 23 '15 at 6:47

• 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 – Inquisitive Jul 23 '15 at 6:38