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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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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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    @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.2012-12-26