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Let $f$ be a holomorphic function on the punctured unit disk. If the imaginary part of $f$ is bounded, is it true that $f$ has a removable singularity at 0?

I see that $|e^{-if}|=e^{Im\;f}$ so $e^{-if}$ is a bounded holomorphic function on the punctured unit disk and it follows that $e^{-if}$ has a removable singularity at 0. The problem is how to say the same for $f$.

cyc
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1 Answers1

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So you have observed that $g=e^{-if}$ is holomorphic on the open disk $D$. Therefore $g'=(-if')e^{-if}=-if'g$ is holomorphic on $D$. So $f'=ig'e^{if}$ is holomorphic on $D$. This implies that $f$ is holomorphic on $D$ (think of the Laurent series and its term by term derivative, for instance).

Julien
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  • Let me see if I understand. So in the Laurent series of $f$ there cannot be terms with negative exponent for otherwise there will be such terms in the Laurent series of $f'$. – cyc Mar 27 '13 at 04:20
  • @learner Exactly. Like power series, you can differentiate Laurent series term by term. And the Lauren series expansions are unique. SO you get $f'$ holomorphic on one side. And the derivative of the Laurent series of $f$ on the other side. They are equal. So you see where the Laurent series came from. It could not have negative exponents in the first place. – Julien Mar 27 '13 at 04:36
  • If the real part of $f$ is bounded instead of the imaginary part, then it seems like the same argument works with $g=e^{f}$ and $f'=g'e^{-f}$. Am I right? – cyc Mar 27 '13 at 04:41
  • Yes, all the same. – Julien Mar 27 '13 at 04:45
  • If you differentiate the Laurent series term by term, what do you do at the $ z^0$ term? Is it possible for $f'$ to have no negative powers but for $f$ to have $1/z$ and no other negative powers? – john w. May 29 '13 at 15:36
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    @johnw. The $z^0=1$ term vanishes. And if there is $1/z$ in $f$, there will be $-1/z^2$ in $f'$. – Julien May 29 '13 at 15:43
  • @julien So the derivative of a function which is meromporphic will have residue zero everywhere? – Eric Auld Dec 31 '13 at 19:35
  • @EricAuld I did not say that... or I don't understand your question, sorry. – Julien Dec 31 '13 at 19:40
  • @julien The way you describe to take the derivative of a Laurent series (get rid of the $z^0$ term, the $1/z$ term becomes $-1/z^2$) seems to imply that the coefficient of $1/z$ in the derivative will be zero. – Eric Auld Dec 31 '13 at 20:39
  • @EricAuld Yes, it is true, as you can differentiate a Laurent series term-by-term. – Julien Dec 31 '13 at 20:48