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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!

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  • $\begingroup$ Why would you have to do something "close to $\;z=2\pi\;$"? There's nothing to worry about that. In fact, the function is analytic in that annulus... Do you want a Laurent series about $\;z=1\;$, or about $\;z=2\pi\;$ or about $\;z=0\;$ ...? $\endgroup$ – DonAntonio Apr 2 '18 at 14:01
  • $\begingroup$ Sorry, I meant the principal part of the Laurent expansion in this annulus $\endgroup$ – JHadamard Apr 2 '18 at 14:07
  • $\begingroup$ It should be around z=0 $\endgroup$ – JHadamard Apr 2 '18 at 14:29
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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 $\;3k-n<0\iff3k<n\ldots\;$

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  • $\begingroup$ Sorry about the stupid questions but I don't see how the second geometric series converges in the annulus since absz>1 $\endgroup$ – JHadamard Apr 2 '18 at 15:45
  • $\begingroup$ @JHadamard I am the stupid, not your question: I completely oversaw that one! I shall edit: it doesn't look too messy. $\endgroup$ – DonAntonio Apr 2 '18 at 15:57
  • $\begingroup$ Thanks! I thought about going about it a bit differently. If I add 1/z^2+1/z the function becomes analytic at 0 and if I add 1/(cos(1)-1)*sum from 0 to infinity of z^(-k) $\endgroup$ – JHadamard Apr 2 '18 at 17:11
  • $\begingroup$ Then that should be the principal part of the Laurent expansion $\endgroup$ – JHadamard Apr 2 '18 at 17:16

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