What kind of singularity have $\frac{1}{1-\cos(z)}$ for z=0?

Question

Set $\frac{1}{1-\cos(z)}$ and $z_0$. I cant identify the kind of singularity since the power series belong in numerator and I have tried with $$\frac{1}{1-\cos(z)}=\frac{1}{1-\cos(z)}\frac{1+\cos(z)}{1+\cos(z)}=\frac{1+\cos(z)}{1-\cos^2(z)}=\frac{1+\cos(z)}{\sin^2(z)}$$ but again have the same problem. And if is a pole how I find the main part for the Laurent series?

Answer 1

$$\frac1{1-\cos z}=\frac1{1-\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-\ldots\right)}=$$

$$\frac1{\frac{z^2}2\left(1-\frac{z^2}{12}+\ldots\right)}=\frac2{z^2}\left(1+\frac{z^2}{12}+\frac{z^4}{144}+\ldots\right)$$

Observe the above already gives us a good look around $\;z=0\;$ and that's why we could omit so many terms in the power series of the denominator.

It is a pole of order two.

Answer 2

near $z=0$ we have $$ \cos z = 1 - \frac12z^2 + O(z^4) $$ hence the pole has order $2$, i.e. the function $$ \frac{z^2}{1-\cos z} $$ has a 'removable' singularity at the origin and approaches the value $2$ as $z \to 0$

Answer 3

Note that

$$ \frac{1}{1-\cos z} = \underbrace{\frac{z^{2}(1+\cos z)}{\sin^{2}z}}_{=:f(z)} \frac{1}{z^{2}} $$

and $f$ has removable singularity at $z=0$. (Just recall that $\sin z/z \to 1$ as $z \to 0$.) This shows that this function has a pole of order 2.