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How can we find the first few terms of the Laurent series of $f(w)={1\over \cos w-1}$ where $w\in \mathbb C, |w|<2\pi$? I am wondering if there is another way maybe exploiting one of the more familiar series expansions instead of brute force formula $c_n={1\over 2\pi i}\int_\gamma {f(w)\over (w-a)^{n+1}}dw$?

Also would it be different if we choose to have $|w|\in (2\pi,4\pi)$ instead?

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2 Answers

For the first question, substitute the MacLaurin series for $\cos w$ and do long division - just keep as many terms as you want in the result.

The answer to the second question is yes. You need to add and subtract the double poles of $f$ at $\pm 2\pi$ $$ f(w) = \left[\frac{1}{\cos w-1} + \frac{2}{(w-2\pi)^2} + \frac{2}{(w+2\pi)^2}\right] - \left[\frac{2}{(w-2\pi)^2} + \frac{2}{(w+2\pi)^2}\right]$$ and expand the binomials in powers of $w$ in the first brackets and powers of $1/w$ in the second.

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Wolphram Alpha gives me the first few terms here:

http://www.wolframalpha.com/input/?i=Laurent+series+of+1%2F%28cos+z+-1%29

I cannot answer your other questions, sorry.

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