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Does there is non constant analytic function from $\{z\in\mathbb{C}:z\neq 0\}$ to $\{z\in\mathbb{C}:|z|>1\}?$ According to me there is no such non constant analytic function because if there is any such function say $f,$ then $f$ can have either a pole or essential singularity at $z=0$. In the case of pole Picard's theorem of meromoprphic function will work and in the case of essential singularity we know that image of any neighbourhood of essential singularity is dense in $\mathbb{C}$, so in both of the cases we get a contradiction. So no such non constant analytic function. Am i right? Please suggest me. Thanks.

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    $\begingroup$ Yes, you're right. $\endgroup$ – Omnomnomnom Jun 20 '16 at 19:17
  • $\begingroup$ picard's theorem for meromorphic theorem says that non constant meromorphic function can omit at most 2 points from its range $\endgroup$ – neelkanth Jun 21 '16 at 2:52
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If so then $1/f$ is bounded. Hence $1/f$ has a removable singularity at the origin, giving a bounded entire function.

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Note that $\mathbb{C}\setminus\overline {\mathbb{D}}$ is conformally equivalent to $\mathbb{D}\setminus\{0\}$, via the map $z \to \dfrac{1}{z}$. So essentially you have to construct a map from $\mathbb{C}\setminus\{0\}$ to $\mathbb{D}\setminus\{0\}$. Suppose the map is $f$. We already have a map, $\mathbb{C}\to\mathbb{C}\setminus\{0\}$ given by $z \to e^z$. Compose this with $f$ to obtain a bounded entire function, which means $f$ is constant.

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  • $\begingroup$ $\mathbb{C} \setminus \overline{\mathbb{D}}$ is conformally equivalent to $\mathbb{D} \setminus \{0\}$, right? $\endgroup$ – Sahiba Arora May 18 '17 at 20:22
  • $\begingroup$ Yes indeed, thanks. $\endgroup$ – Hmm. May 18 '17 at 20:38

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