# If $f$ has a essential singularity at $P$, then for $(z-P)^m f(z)$ has also an essential singularity.

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?

• You should look at the definition of ess sing again. It is not quite as simple as you state it. – H. H. Rugh Nov 1 '16 at 13:58
• 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. – phy_math Nov 3 '16 at 3:49

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.