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Let f and g be two holomorphic functions on a domain $\Omega$. Suppose that $\frac{f}{g}$ is always finite (while g can be zero at some points). Is it true that then $\frac{f}{g}$ is holomorphic? Actually I do not know any example of 2 such functions, except of course the trivial case $f=cg$ where c is a constant. It will be thus very helpful if someone can provide nontrivial examples. Thanks in advance!

Edit: finally, this is standard thing about removable singularities. Nothing more needed! Thanks!

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  • $\begingroup$ An easy example: $f(z)=z^{n+1}$, $g(z)=z^n$ $\endgroup$
    – S -
    Jul 29, 2014 at 9:22
  • $\begingroup$ You might like to look at this Wikipedia page. $\endgroup$
    – user149874
    Jul 29, 2014 at 9:24
  • $\begingroup$ @Adolfo Thanks for your comment! About that easy example, well I'm looking for examples different from that kind. To produce examples of that kind, we take h holomorphic and we write h= (hf)/f :). I want to see more nontrival examples, where f and g don't have easily seen common factors. $\endgroup$ Jul 29, 2014 at 9:30
  • $\begingroup$ How about $g=e^z$? There are entire functions. $\endgroup$ Jul 29, 2014 at 9:32
  • $\begingroup$ @Jez Thanks. I see your point, so it seems that the answer is affirmative. $\endgroup$ Jul 29, 2014 at 9:33

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Answer: yes. I am going to formalize where you said "$f/g$ is always finite" to mean: "for any $P \in \Omega$, if $g$ has a zero at $P$, then $f$ has a zero of at least the same order."

Recall that we define the order of a zero to be smallest $n \geq 0$ such that $f^{(n)}(P) \neq 0$ but $f^{(n-1)}(P) = 0$. That this exists (when the function is not identically the zero function) is a consequence of holomorphicity, which is obvious if you already know that holomorphic functions are analytic.

Given that $f$ and $g$ are holomorphic, this is the only hypothesis needed to ensure that $f/g$ admits a holomorphic continuation to a domain including $P$. Moreover, this condition is necessary. This is seen easily by Taylor expanding at $P$.

In particular, if a continuous extension exists, it is automatically holomorphic.

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  • $\begingroup$ thanks, I get it. Being a physics major, sometimes I forget to formalize things properly. Another naive question, what about real functions (smooth but not analytic for example)? From your answer, it seems to me that holomorphicity=analyticity is a key ingredient. $\endgroup$ Jul 29, 2014 at 10:50
  • $\begingroup$ i don't know how to formalize the question for such a function, e.g. consider a bump function; what is its order of vanishing at the origin? $\endgroup$
    – hunter
    Jul 29, 2014 at 11:27

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