Find the first three nonzero terms in the Laurent series expansion of $f$ in a region

Question

$f(z) = 1/(1-cosz)$

Region is $0 < |z| < 2pi$

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My work:

$f(z) = 1/(1-cos(z-2pi))$

$1-cos(z-2pi) = (z-2pi)^2/2! - (z-2pi)^4/4! + (z-2pi)^6/6! - ....$ (*)

--> $1/$(*) $= 2/(z-2pi)^2 + 1/6 + (z-2pi)^2/120 + ...$

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I don't know what the answer is, but I doubt it's this. Any idea on how to go about doing it?

Answer 1

\begin{align} f(z) &= \dfrac{1}{1-\cos z} \\ &= \dfrac{1}{\frac{1}{2!}z^2-\frac{1}{4!}z^4+\frac{1}{6!}z^6+\cdots} \\ &= \frac{2}{z^2}\left(\dfrac{1}{1-\frac{1}{12}z^2+\frac{1}{360}z^4+\cdots}\right) \\ &= \frac{2}{z^2}\left(1+\frac{1}{12}z^2+\frac{1}{240}z^4+\cdots\right) \\ &= \frac{2}{z^2}+\dfrac{1}{6}+\frac{1}{120}z^2+\cdots \end{align}