# $f$ is holomorphic in $B(z_0,r)\setminus\{z_0\}$ and does not except real values. Then $z_0$ is a removable singularity

$$f$$ is holomorphic at $$B(z_0,r)\setminus\{z_0\}$$ and $$f$$ doesn't except real values - i.e $$f(z)\notin\mathbb{R}$$ for all $$z\in \mathbb{R}$$. Then $$z_0$$ is a removable singularity point ($$f$$ can be extended holomorphically in $$z_0$$).

Well I tried to use Riemann theorem, and show that $$\exists 0 such that $$f$$ is bounded in $$B(z_0,r)$$, but didn't succeed to do so. Formerly I solved a similar question which demand that $$\Re(f)>0$$ and then by defining $$e^{-f(z)}$$ which is holomorphic and bounded, which by taking $$log$$ holomorphic branch promises $$f$$ is holomorphic. Is there any manipulation or composition I may make to $$f$$, to get a similar results?

I also tried to assume that $$f$$ is not bounded, so in $$B(z_0,r)$$ one may find $$z$$ such that $$|f(z)|$$ is arbitrary big. However, is there any kind of intermediate value principle which assures that $$f$$ must "cross" the real line in case $$|f(z)|$$ is not bounded in $$B(z_0,r)$$?

• Maybe you can just solve it in the same way you did the $\mathbb{R} > 0$ case just taking $if$ or $-if$. Notice that since the function doesn't take real values, the imaginary part of $f$ must be either positive or negative. – astrobarrel Jul 2 at 16:22
• @astrobarrel Imaginary part must be positive or negative since $f$ preserves connectivity? – dan Jul 2 at 16:27
• Just for the meanvaue theorem. Write $f$ as $f_1+if_2$. $f_2$ is continuous and real. So, assume for a moment that there are two points $z_0, \, z_1$ such that $f_2(z_0)<0<f_2(z_1)$. Take a continuous path $\gamma$ that goes from $z_0$ to $z_1$ and take the composition $f(\gamma)$. Now the meanvalue theorem takes care of it. – astrobarrel Jul 2 at 16:55

EDITED: If $$f$$ has a pole at $$z_0$$, $$1/f$$ has a zero there, and by the Open Mapping Theorem $$1/f$$ would take all values in some interval near $$0$$.

If $$f$$ has an essential singularity at $$z_0$$, Picard says it can omit at most one value near $$z_0$$.

Removable is all that's left.

EDIT If you don't want to use the heavy artillery of Picard, note that if $$f$$ takes no real values, since $$B(z_0,r)$$ is connected the values it does take are either in the upper or lower half plane.

• I'm sorry, can you explain why does $1/f$ takes all real values near zero by the open mapping theorem? – astrobarrel Jul 2 at 17:00
• Maybe I wasn't as clear as I should have been. I meant "all the real values that are in some neighbourhood of $0$". After removing the removable singularity of $1/f$ at $z_0$, the resulting non-constant analytic function maps a neighbourhood of $z_0$ to a neighbourhood of $0$, and any neighbourhood of $0$ contains some real interval $(-\epsilon, \epsilon)$. – Robert Israel Jul 3 at 2:18
• I should have understood that anyway, it was obviuous. Thanks for the clarification. – astrobarrel Jul 3 at 16:58

I am not able to comment in Robert Israels answer however could a possible answer as to why f has no essential singularity come down to the fact that:

Assume f has an essential singularity at $$z_0$$. Then due to the Casorati-Weierstrass Theorem, if V is a neighbourhood around the singularity $$z_0$$ in $$𝐵(𝑧_0,𝑟)∖{𝑧_0}$$ then $$f(V\setminus{z_0})$$ is dense in $$\mathbb{C}.$$ This means that any element in $$\mathbb{C}$$ can be approximated arbitrarily be elements of $$f(V\setminus{z_0})$$. Since $$\mathbb{R}$$ is in $$\mathbb{C}$$ then it also takes on real values. Which results in a contradiction and thus f does not have an essential singularity at $${z_0}$$.

Would this work as an explanation as to why f has no essential singularity at $$z_0$$?

• I don't think so. The fact that $f$ approximates real values doesn't imply itself that $f$ assumes real values. Am I wrong? – astrobarrel Jul 3 at 16:57