Let $f(z) = e^{\frac{1}{1-\cos z}}$.
I want to show that the Laurent's expansion of $f$ near $z = 0$ has infinitely many positive and negative powers of $z$.
For this, I have shown that $z = 0$ is an essential singularity therefore the Laurent expansion of $f$ will have infinitely many negative powers.
I am not able to show that it has infinitely many positive powers.
I tried using the expansion of $\cos z$. The process is becoming really complicated.
Help, please
Best Answer
We use the theorem: if $g(z)$ has a non-removable singularity at $a$, then $e^{g(z)}$ is an essential singularity at $a$.
Infinitely many negative terms: Let $g = \frac{1}{1-\cos{z}}$. When $z = 0$, $g$ has a pole, so $e^g = f$ has an essential singularity.
Infinitely many positive terms: We show $f(1/z)$ has an essential singularity at $0$, where $f(1/z) = \exp(\frac{1}{1 - \cos(1/z)})$. $\frac{1}{1 - \cos(1/z)}$ has an essential singularity at $0$ (limit does not exist). So again $f(1/z)$ has an essential singularity at $0$.