# Evaluate the order of the pole in $1/(1-\cos(z))^2$

I'm trying to determine the order of the pole in the complex expression

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

I have determined the pole to be $$z=2\pi n, n\in \mathbb{Z}$$.

However, when I use the equation $$\lim\limits_{z\rightarrow 2\pi n}[(z-2\pi n)^k f(z)]$$ with $$k=1$$, it equals $$\frac{0}{1}=0$$ or that the function is analytical in the neighborhood. I have used L'Hôpital's rule repeatedly to obtain this result. I checked my answer with Wolfram Alpha, and it's supposed to have a pole of order $$4$$ and $$z=2\pi n$$. Where am I going wrong?

HINT:

Note that $$1-\cos(z)=2\sin^2(z/2)$$. Thus, $$\left(1-\cos(z)\right)^2=4\sin^4(z/2)$$

Can you proceed now?

• Thank you so much :-) May 16, 2021 at 15:31
• You're welcome. My pleasure. May 16, 2021 at 16:06

If a function $$g$$ has a pole of order $$k$$ then $$g^2$$ will have a pole of order $$2k$$. Using your method, we have by L'Hopital, $$\lim_{z\to2\pi n}\frac{z-2\pi n}{1-\cos z}=\lim_{z\to2\pi n}\frac1{\sin z}=\infty$$ but $$\lim_{z\to2\pi n}\frac{(z-2\pi n)^2}{1-\cos z}=\lim_{z\to2\pi n}\frac{2(z-2\pi n)}{\sin z}=\lim_{z\to2\pi n}\frac2{\cos z}=2$$ so $$1/(1-\cos z)$$ has a pole of order $$2$$. Therefore, $$f(z)=1/(1-\cos z)^2$$ has a pole of order $$4$$.

In your attempt, the limit with $$k=1$$ is in fact of the form $$1/0$$ not $$0/1$$, so $$f$$ is not analytic as you claim.

• I got the limit of 0/1 and not 1/0, because I assumed that 0/1 was undefined as it approaches infinity from above and - infinity from below. Therefore I used L'Hôpital again. Does the sign of the infinity not matter in this instance? Aside from this, I found your answer very clear, thanks! May 16, 2021 at 15:23
• When $k=1$ we just need L'Hopital once as shown below: $$\lim_{z\to2\pi n}\frac{z-2\pi n}{(1-\cos z)^2}=\lim_{z\to2\pi n}\frac1{-2\sin z(1-\cos z)}=\frac10$$ May 16, 2021 at 15:25
• But doesn't this only evaluate as $\infty$ when taken as the upper limit? May 16, 2021 at 15:28
• It doesn't matter as the limit is indeterminate so the pole cannot be of order $\le1$. May 16, 2021 at 15:29
• Ah, thank you so much May 16, 2021 at 15:31