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Suppose $g(z)$ has an isolated singularity at $z=z_0$ and $|\Re[g(z)]| \ge M>0$ for all $z \in \mathbb C-\{z_0\}$. What is the type of singularity of $g$ at $z_0$?

I have a guess it is removable but I could not argue why it can not be a pole though. To rule out essential, I argue with Casorati-Weierstrass theorem.


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How do you define $\text{Reg} z$, and what is $M$? –  Fly by Night Dec 28 '12 at 0:26
    
That's the real part of $g$ and it takes the value outside the strip of $-M$ to $M$ –  Deepak Dec 28 '12 at 0:29
    
I've edited the LaTeX. Now, what is the question? You have given an assumption, but what would you like us to do with that assumption? –  Fly by Night Dec 28 '12 at 0:45
    
@FlybyNight, I am sorry, I need to figure out what kind of singularity does $g$ have at $z_0$? –  Deepak Dec 28 '12 at 0:46
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1 Answer

As a counter example, consider the following function:

$$g(z) = \frac{z^2}{z(z-1)} \, . $$

This function has a removable singularity at $z=0$. Consider the domain $\mathbb{C}\backslash\{0\}$. You claim that there exists an $M>0$ such that $|\Re[g(z)]| \ge M > 0$ for all $z \in \mathbb{C}\backslash\{0\}$. However, this is not true.

We see that, assuming $z \neq 1$, $\Re[g(z)] = 0 \iff x^2-x+y^2=0 \iff |z-1/2|=1/2$. So there is no $M > 0$ for which $\Re[g(z)] \ge M > 0$ for all $z \in \mathbb{C}\backslash\{0\}$.

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I just realized that if we consider function $\frac {1}{g(z)}$, then that function will be analytic using Riemann Theorem on removable singularity, and is entire of course. Then Liouville's applied to $\frac{1}{g(z)}$ going to tell you the function is constant and hence $g(z)$ is a constant function. That means the singularity is removable. I don't know if I am right on this? You example rocks, I can not see a point to doubt on that. It seems the way I did also make sense. Would you plase comment further? –  Deepak Jan 8 '13 at 17:30
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