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$f$ is analytic in $\Omega-{\{z_o}\}$, where $\Omega$ is a domain in $\mathbb C$. If $\text{Im}f(z)>= -B$ for all $z\in \mathbb C$. I am trying to figure out nature of singularity $f$ have at ${z_0}$.

Obviously, Casorati-Weirestrass comes in to play. We can conclude it can not have essential singularity at ${z_0}$. I can not see $f$ bounded near $z_0$ and neither I see $f(z)$ going to $\infty$ as $z\rightarrow z_0$. I am missing something here. Please help.

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up vote 1 down vote accepted

As you say, $f$ can neither be an essential singularity nor a pole. But it can be bounded near $z_0$. For example: if $\Omega$ is the upper half plane, $z_0$ is any point in $\Omega$, and $f$ is the identity function.

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$f$ is defined in some domain $\Omega$ minus one point $z_0$, not necessarily on $\mathbb{C}$ minus $z_0$... – Kalim Dec 26 '12 at 2:24
@Malik: Oops, I missed that. Fixed. – Ted Dec 26 '12 at 3:19
how do you rule out that it cannot be a pole – Koushik Dec 26 '12 at 5:24
@K.Ghosh In the neighborhood of a pole, a meromorphic function takes all values in a neighborhood of infinity (i.e. some region $|z|>R$), so the range could not lie in a half plane. – Ted Dec 26 '12 at 6:38
@Deepak Look at the prototypical function with a pole of order $n$ at 0: $f(z) = z^{-n}$. In a neighborhood of 0, say $|z|<\epsilon$, its range is $|z|>1/\epsilon^n$, which is the complement of a closed disc, which does not contain any half-plane. In general, a function with a pole of order $n$ at 0 will have a series expansion $cz^{-n}$ plus higher powers of $z$. In a small neighborhood of 0, the behavior of $f$ will be similar to $z^{-n}$, and the range will contain (though not be equal to) some $|z|>R$. As long as the range contains $|z|>R$ it cannot be contained in a half-plane. – Ted Dec 26 '12 at 16:04

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