Extension of essential singularity.

Question

I need to proof that if a function $f$ has an essential singularity at $z_{0}$ then can be expressed as a sum of the form:

$f(z)=\sum_{-\infty}^{+\infty}a_{n}(z-z_{0})^n$

I really don't have idea how to do this problem.

Answer 1

More precisely, if $f$ has an essential singularity at $z=z_0$, then $f$ is expressed as a Laurent series around $z_0$: $$ f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n, $$ with infinitely many non-vanishing terms with negative index.

Fact I. If $f$ is analytic in $A=\{z\in\mathbb C: 0<|z-z_0|<r\}$, then $f$ can be expressed as a Laurent series around $z_0$: $$ f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n, $$ which converges (locally uniformly) for every $z\in A$.

Fact II. If $f$ is analytic in $A=\{z\in\mathbb C: 0<|z-z_0|<r\}$, and $z_0$ is a removable singularity then $a_n=0$, for all $n<0$. If $z_0$ is a pole, then there exists a $k\in\mathbb N$, such that $a_n=0$, wherever $-n>k$ - $k$ is the degree of the pole. Finally, if $z_0$ is an essential singularity, then $a_n\ne 0$, for infinitely many negative $n$'s.