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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

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  • $\begingroup$ Do you know how to divide power series? $\endgroup$ Nov 4, 2016 at 1:46
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    $\begingroup$ 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. $\endgroup$
    – wgrenard
    Nov 4, 2016 at 1:54
  • $\begingroup$ Yes, there is removable singularity. $\endgroup$ Nov 4, 2016 at 1:56
  • $\begingroup$ @JackyChong Is it done just term by term? $\endgroup$
    – Quality
    Nov 4, 2016 at 2:20

1 Answer 1

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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.

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