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

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.

$$g(z) = \frac{z^2}{z(z-1)} \, .$$
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 $z \in \mathbb{C}\backslash\{0\}$.
• 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? Commented Jan 8, 2013 at 17:30