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Let $f$ be an entire function and $L$ a line in $\mathbb{C}$ such that $f(\mathbb{C})\cap L=\emptyset$. Show that $f$ is constant function.

If $f$ is not constant then $f(\mathbb{C})$ is dense set in $\mathbb{C}$, but how can I use that line does not intersect the image set?

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  • $\begingroup$ Little Picard would work; it's not clear what you have to work with since you haven't given any context. $\endgroup$ – Chappers Feb 25 '17 at 16:31
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Hint Show that $f(\mathbb{C})$ is (path) connected.

Since $f(\mathbb C)$ doesn't intersect the line, it is included in one of the half planes defined by $L$.

Next, pick some $w$ in the other half plane, and show that $$g(z)=\frac{1}{f(z)-w}$$ is entire and bounded.

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  • $\begingroup$ @ N.S how to show $g(z)$ is bounded?? $\endgroup$ – Eklavya Feb 25 '17 at 16:42
  • $\begingroup$ @Eklavya Set $R=$ distance from $w$ to $L$. Then, $B_R(w)$ is not meeting $L$, and hence $|f(z)-w| \geq R$. $\endgroup$ – N. S. Feb 25 '17 at 16:48
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Wlog. $L$ is the $x$-axis and $f(\Bbb C)\subseteq \Bbb H$. Note that you can map $\Bbb H$ to $\Bbb D$.

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As $f$ is entire and non-constant, $f$ takes all values with atmost one exception possibly (Little Picard's theorem). But $f(\mathbb C)\cap L=\emptyset$ implies $f$ skip infinite points on $\mathbb C$, so $f$ must be constant.

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