3
$\begingroup$

$f$ 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 $\exists 0<r'\le r$ 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 $B(z_0,r)$?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – astrobarrel Jul 2 at 16:22
  • $\begingroup$ @astrobarrel Imaginary part must be positive or negative since $f$ preserves connectivity? $\endgroup$ – dan Jul 2 at 16:27
  • 1
    $\begingroup$ 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. $\endgroup$ – astrobarrel Jul 2 at 16:55
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I'm sorry, can you explain why does $1/f$ takes all real values near zero by the open mapping theorem? $\endgroup$ – astrobarrel Jul 2 at 17:00
  • $\begingroup$ 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 $(-\epsilon, \epsilon)$. $\endgroup$ – Robert Israel Jul 3 at 2:18
  • $\begingroup$ I should have understood that anyway, it was obviuous. Thanks for the clarification. $\endgroup$ – astrobarrel Jul 3 at 16:58
0
$\begingroup$

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 $z_0$?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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