Let $f$ be an analytic function in the annulus $ \{ 0<|z| Is it true that the Laurent series expansion of $f$ about $0$ has a finite number of negative coefficients? It came to my mind since $0=r=\limsup_{n\rightarrow\ \infty}|a_{n}|^{1/n}$ and therefore $|a_{n}|$ must be $0$ at some point. Am I correct?
Laurent expansion of an analytic function in a neighborhood of 0
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complex-analysis
laurent-series
1 Answers
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First a Laurent expansion is defined in an annulus and thus the statement: "Laurent expansion about $0$" makes little sense. If you are refering to the Laulernt expansion at $\left\{0<\left|z\right| If $f$ has a removable singularity at $0$, there won't be any negative powers. If $f$ has a pole there, there will be finitely many negative powers. If $f$ has an essential singularity there, there will be infinitely many negative powers. The coeffecients may very well be negative. Oh and your proof is incorrect. $$\lim\sup\left|a_n\right|^{\frac1n}=0$$
doesn't imply $a_n=0$ for large $n$.
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0For example, the Laurent series for $\sin(1/z)$ has infinitely many negative powers. – 2012-12-24
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0@GEdgar or even $e^{1/z}$ – 2012-12-24