Basic example of classifies singularities in complex analysis

Question

I want to classify the singularity at $z=0$ of the function

$$f(z)=\frac{z^2}{1-\cos z}$$

My book says, the answer is pole.

I know that the options are either, it could be a isolated removable singularity, a pole, or an essential singularity.

But here is what I thought; the limit as $z$ goes to $ 0$ is clearly $2$ via L'Hospital rule, but my notes also say that p is a pole if the limit as we approach the absolute value of the function is infinity. So I am having trouble seeing why this is a pole.

On the other hand, since the limit is 2, there will certainly exist a $r>0$ such that $|f(z)| \le 2$ for all $z$ in the punctured disk around it.

So that leads me to believe that we are dealing with a removable singularity..

Is that correct? Or the book correct and I am missing something cruicial?

Thanks

Answer 1

Remember the power series expansion for $\cos(z)$:

$$\cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}.$$

So plug into your function:

$$f(z)=\frac{z^2}{1-\cos(z)}=\frac{z^2}{1-(1-\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots)}=\frac{1}{\frac{1}{2!}-\frac{z^2}{4!}+\frac{z^4}{6!}+\cdots}$$

So it is easier to see now that $z=0$ is not a pole of $f(z)$ but a removable singularity with $\lim_{z\to0}f(z)=\frac{1}{1/2!}=2$, as you calculated.