# Basic example of classifies singularities in complex analysis

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

• Do you know how to divide power series? Nov 4, 2016 at 1:46
• I agree that it's a removable singularity. If you take a look at the Laurent expansion around $z = 0$ this is obvious because there are no terms with a power of $z$ in the denominator. wolframalpha.com/input/?i=laurent+series+z%5E2%2F(1+-+cos(z)) Also, as you noted, it is also pretty obvious that the function is bounded around zero, so it can't be a pole. Nov 4, 2016 at 1:54
• Yes, there is removable singularity. Nov 4, 2016 at 1:56
• @JackyChong Is it done just term by term? Nov 4, 2016 at 2:20

Remember the power series expansion for $\cos(z)$:
$$\cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}.$$
$$f(z)=\frac{z^2}{1-\cos(z)}=\frac{z^2}{1-(1-\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots)}=\frac{1}{\frac{1}{2!}-\frac{z^2}{4!}+\frac{z^4}{6!}+\cdots}$$
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.