# How to find the principal part of the Laurent expansion of $\frac{1}{(\operatorname{cos}(z)-1)^2}$

I'm confused how to find the principal part of the Laurent expansion of this function?

$\frac{1}{(\operatorname{cos}(z)-1)^2}$ in $0<|z|<\pi$

If you factor out negative 1 then can you expand it using binomial theorem? By doing this you find no negative powers? I would very much appreciate being pointed in the correct direction!

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Could you explain a bit more about your idea of "factoring out negative 1"? It's not clear to me what that means. –  Henning Makholm Nov 9 '11 at 21:46
$(1-\operatorname{cos}(z))^{-2}$ is the same as the function stated above, can we not use binomial theorem to expand this? as it is in the form $(1+x)^{-n}$ –  Freeman Nov 9 '11 at 21:49
@LHS You don't want to do that, because the binomial theorem works for small x, and cos is not small near 0. –  Phira Nov 9 '11 at 21:53
Perhaps, but is looks like this will lead you into a little maze of twisty Chebyshev polynomials. What I would do is notice that $\cos(z)-1=-2\sin^2(z/2)$ to get rid of the subtraction. –  Henning Makholm Nov 9 '11 at 21:57

Laurent series expansion at what point? The function is holomorphic everywhere in $\mathbb{C}\backslash\{2k\pi, k\in\mathbb{Z}\}$. This means that for any $z_0\neq 2 k \pi$, the Laurent series expansion at $z_0$ will have a zero principal part.
As for the Laurent series expansion at any of the singularities, consider for example the expansion at $z_0=0$. Since for $|z|\ll 1$, $\cos z-1 = -\frac{1}{2}z^2 + \frac{1}{24}z^4+O(z^6)$ we have $$\frac{1}{(\cos z-1)^2} = \frac{4}{z^4}\left(1-\frac{1}{12} z^2 + O(z^4)\right)^{-2} = \frac{4}{z^4} + \frac{2}{3 z^2} + O(1)$$ So the principal part of the Laurent series expansion at $z=0$ is $\frac{4}{z^4} + \frac{2}{3 z^2}$.