I am trying to find the principal part of the Laurent expansion of $$\frac{1}{(\cos(z)-1)(z-1)}$$ in the annulus $1<|z|<2\pi$. I tried doing it by expanding $\cos(z)-1$ in a neighbourhood of $1$ and then rewrite the other part as a geometric series. The problem is I don't know what to do close to $|z|=2\pi$.
Help would be much appreciated!
Well, we have
$$\cos z-1=\sum_{n=0}^\infty\frac{(-1)^nz^{2n}}{(2n)!}-1=\sum_{n=1}^\infty\frac{(-1)^nz^{2n}}{(2n)!}=$$
$$=-\frac{z^2}2+\frac{z^4}{24}-\ldots=-\frac{z^2}2\left(1-\frac{z^2}{12}+\ldots\right)\implies$$
$$\frac1{(\cos z-1)(z-1)}=-\frac1{\frac{z^2}2\left(1-\frac{z^2}{12}+\ldots\right)}\cdot\left(\frac1z\cdot\frac1{1-\frac1z}\right)=$$
$$=-\frac2{z^3}\left(1+\frac{z^2}{12}+\frac{z^4}{144}+\ldots\right)\left(1+\frac1z+\frac1{z^2}+\ldots\right)=$$
$$=-\frac2{z^3}\sum_{n=0}^\infty\frac{z^{2n}}{12^n}\cdot\sum_{n=0}^\infty\frac1{z^n}=-2\sum_{n=0}^\infty\sum_{k=0}^n\frac{z^{2k}}{12^k}\frac1{z^{n-k+3}}=-2\sum_{n=0}^\infty\sum_{k=0}^n\frac{z^{3k-n-3}}{12^k}=$$
$$=-2\sum_{n=0}^\infty z^{-n}\sum_{k=0}^n\frac{z^{3k}}{12^k}$$
You can now try to get the explicit principal part from the above, or at least write down some elements of it, for example: for some fixed $\;n\in\Bbb N\;$ ,we get
$$-2\,z^{-n}\left(1+\frac{z^3}{12}+\frac{z^6}{12^2}+\ldots+\frac{z^{3n}}{12^n}\right)$$
The above will yield summands of the principal part iff $\;3k-n<0\iff3k<n\ldots\;$