Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f$ 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.

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
$f$ is defined in some domain $\Omega$ minus one point $z_0$, not necessarily on $\mathbb{C}$ minus $z_0$... –  Malik Younsi 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 –  K.Ghosh Dec 26 '12 at 5:24
1  
@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
1  
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.