# Singularities of $f(z)=\frac{z^4}{\cos(z^2)-1}$

I'm having difficulties classifying the isolated singularities of the function

$$f(z)=\frac{z^4}{\cos(z^2)-1}.$$

The function $f$ is undefined when the denominator equal $0$, that is

$$\cos(z^2)-1=0 \iff \cos(z^2)=1 \iff z^2 = 2\pi n \iff z = \pm\sqrt{2\pi n}, \; n\in \mathbb{Z}.$$

So $f(z)$ has isolated singularities at every point $z=\pm\sqrt{2\pi n}, \; n\in \mathbb{Z}$.

Obviously I can't take the Laurent series of $f$ around every singularity to determine it's nature so how do I continue from here?

• From the taylor expansion of $\cos(z)$ it is natural to guess that all singularities for $n\neq 0$ are poles of order $4$, whereas the singularity at $0$ is removable. Now just prove this by checking the definition. Of course you need only to check it in the two above mentioned cases, once for $n=0$ and once for a general $n\neq 0$. – b00n heT Jul 1 '16 at 15:43
• @b00nheT The definition you are referring to is: "$z_0$ is a removable singularity if and only if the $\lim_{z \to z_0}f(z)$ exists and is finite". Is that correct? – glpsx Jul 1 '16 at 15:58
• Yep. You need to compute some limits to solve it the way I suggest – b00n heT Jul 1 '16 at 15:59
• check egreg's answer to see my comment with the explicit calculations – b00n heT Jul 1 '16 at 16:17

## 1 Answer

Consider the easier function $g(z)=\dfrac{z^2}{\cos z-1}$, so $f(z)=g(z^2)$.

For $g$, the singularity at the origin is removable, because $$\lim_{z\to0}g(z)=-2$$ In order to examine the singularities at $2k\pi$ for $k$ an integer, $k\ne0$, consider $$\lim_{z\to2k\pi}\frac{z^2(z-2k\pi)^2}{\cos z-1}= \lim_{w\to0}\frac{(w+2k\pi)^2w^2}{\cos w-1}=-8k^2\pi^2$$ with the obvious substitution $z=w+2k\pi$. Thus these points are poles of order $2$ (with residue $0$).

If $g$ has a pole of order $2$ at $z_0$, then $f(z)=g(z^2)$ has a pole of order $4$ at $\sqrt{z_0}$ (either determination): just substitute in the Laurent expansion.

• Thank you for your detailed and clear explanation. It is extremely helpful and understandable. The "trick" to show that $f$ has a removable singularity at $0$ is amazing. – glpsx Jul 1 '16 at 16:21