How would one show that if $\displaystyle \lim_{z \to 0} z \exp(f(z))$ exists, where $f: \mathbb{C}^\ast \longrightarrow \mathbb{C}$ is holomorphic, then it must be zero? Is this even true?
$\lim_{z \to 0} z \exp(f(z))$ for holomorphic $f$?
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0It means that $\exp(f(z))$ cannot have a simple pole. – 2012-05-21
2 Answers
If $f$ is holomorphic on $\mathbb{C}^*$, then it has either a removable singularity, a pole, or an essential singularity at $0$.
If it has a removable singularity, $\lim_{z \to 0} f(z)$ exists, which means the limit you're looking for is $0$. In the other two cases, you should be able to show that the limit doesn't exist (in the case where it has an essential singularity, you might need a big stick like Picard's Theorem to do this).
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0Yes, $\mathbb C^*$ is notation from another branch of mathematics. The OP should better have said it more explicitly, so that complex analysts would understand it. – 2012-05-21
As noted above, think of the singularity type of $f$. If it is removable, you are done. If it is polar, write $f(z)=h(z)/z^n$ and expand the Taylor series for $z*Exp(f(z))$ to see that it has an essential singularity near 0, so there is no limit. If $f$ has an essential singularity, by Big Picard it assumes all but one complex value infinitely often near $0$; its exponential also has this property, so it is has an essential singularity as well.