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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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    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.2011-11-09

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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 $\frac{4}{z^4} + \frac{2}{3 z^2}$.

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    How about in the region I specified?2011-11-10