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Is there a function $f$ that is holomorphic on $\mathbb{C}-\mathbb{Z} $ and maps into or onto $\mathbb{C}-\mathbb{R}$ ? Into or onto $\mathbb{C}-\mathbb{R}^{+}\cup\{ {0} \}$?

All I have been able to deduce is that $f$ must be meromorphic (and certainly not entire), or else it would contain an essential singularity, and the existence of $f$ would contradict Picard's Great Theorem.

Alternatively, if $g$ is holomorphic on $\mathbb{C}-\mathbb{Z}$ and maps onto $\mathbb{C}-S$, can we conclude that $|S|\leq |\mathbb{Z}|$ ? Does the continuum hypothesis come into play?

EDIT: Exclude constant functions.

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    $\begingroup$ Image of a connected set is connected so it definitely cannot be onto $\endgroup$ – happymath Jul 23 '16 at 10:52
  • $\begingroup$ Adding $\{0\}$ doesn't change anything since the image of $f$ has to be open. $\endgroup$ – Tim B. Jul 23 '16 at 10:58
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As you already noticed, Picard and tha fact that more than one value is left out tells us that the singularities at points in $\Bbb Z\cup \{\infty\}$ are non-essential. On the other hand, if $n\in\Bbb Z$ is a pole of order $k>0$, then the image under $f$ of a sufficently small circle around $n$ will wind $k$ times around $\infty$ and thus intersect the real line. Hence all singularities are removable, and then $f$ is entire. But then $f$ must be constant.

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  • $\begingroup$ Is it possible for this method to be generalised? In other words, if $f:\mathbb{C}-\mathbb{Z}\rightarrow\mathbb{C}-S$ is holomorphic, can we conclude that $|S|<|\mathbb{R}|$? Sorry to trouble you. $\endgroup$ – Yon Teh Jul 23 '16 at 12:13
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There is no such function:

First, we observe, that $f$ maps interely into $\{z\in\mathbb C: Im(z)>0\}$ or $\{z\in\mathbb C: Im(z)<0\}$, since its image is open (and therefore $f$ has to ommit the value $0$) and path-connected. We assume that $f$ maps into $\{z\in\mathbb C: Im(z)>0\}$.

As you already mentioned, $f$ can't have an essential singularity. But $f$ can't have a pole either: If w.l.o.g. $f$ had a pole in $0$, consider its Laurent expansion $\sum_{n=-k}^\infty a_n z^n$. Then there is $\epsilon>0$ such that $|a_k z^{-k}| > 2|\sum_{n=-k+1}^\infty a_n z^n|$ if $|z|<\epsilon$.

Now take a $k-$th root of $\theta$ of $\frac{i}{a_k}$. Let $c>0$ be small enough that $|c\theta|<\epsilon$. Then $a_k(c\theta)^{-k} = \frac{-i}{c^k}$. Note that $f(c\theta)$ must then have negative imaginary part which contradicts our assumption.

So $f$ must be entire which is a contradiction aswell.

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