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Prove that if a function $f$ is holomorphic on $D(P,r) \setminus \{P\}$ and has an essential singularity at $P$, then for any integer $m$ the function $(z-P)^m f(z)$ has an essential singularity at $P$.


My strategy, $f$ has an essential singularity at $P$ means $\lim_{z \rightarrow P}f(z)$ is not well defined. To show $(z-P)^m f(z)$ has an essential singularity at $P$, i want to show $\lim_{z \rightarrow \infty} (z-P)^m f(z)$ is also not well defined.

But i am not sure if $\lim_{z \rightarrow P}f(z)$ is not well defined, then $\lim_{z \rightarrow \infty} (z-P)^m f(z)$ for any integer $m$.

How can i prove this?

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  • $\begingroup$ You should look at the definition of ess sing again. It is not quite as simple as you state it. $\endgroup$ – H. H. Rugh Nov 1 '16 at 13:58
  • $\begingroup$ As far as I known, essential singularity is a singularity which is not a pole $\lim_{z \rightarrow P} |f(z)| = \infty$ and removable singularity, $ |f(x)|\leq M$, for some $M>0$. I tried to solve above problem by ill-definedness limit of $(z-P)^m f(z)$, but i am not sure of the last statement that i wrote above question. $\endgroup$ – phy_math Nov 3 '16 at 3:49
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At an essential singularity, the Laurent series of $f$ centred at $P$ has infinitely many non-zero negative terms $$ f = \sum_{j=-\infty}^\infty a_j (z-P)^j \qquad (z - P)^m f = \sum_{j=-\infty}^\infty a_{j-m} (z-P)^j $$so $(z - P)^m f$ has infinitely many non-zero negative terms at $P$ as well, so it has an essential singularity.

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