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Compute the principal part of the Laurent series of $\cot(\pi z)$ on $1<|z|<2$.

EDIT: After using either of the approaches below, we get that the principal part is equal to

$$ \frac 1 \pi \left( \frac 3 z + \sum_{n=1}^\infty \frac 2 {z^{2n+1}}\right)$$

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    $\begingroup$ Do you think that logarithm is well-defined around the whole circle? $\endgroup$ – GEdgar Jun 1 '15 at 2:20
  • $\begingroup$ oh, good point. $\endgroup$ – Ashley Jun 1 '15 at 2:22
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Recalling that: $$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right)\tag{1}$$ and considering $\frac{d}{dz}\log(\cdot)$ of both sides we have: $$ -\frac{1}{z}+\pi\cot(\pi z) = \sum_{n\geq 1}\frac{2z}{z^2-n^2}\tag{2} $$ or: $$ \cot(\pi z)=\frac{1}{\pi z}+\frac{2}{\pi}\sum_{n\geq 1}\left(\frac{1}{z-n}+\frac{1}{z+n}\right).\tag{3}$$ Can you finish from there?

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    $\begingroup$ yes. thanks much! $\endgroup$ – Ashley Jun 1 '15 at 18:10
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Another approach: In $\{|z|<2\},$ $\cot \pi z$ has only three singularities, namely simple poles at $0,1,-1.$ If we let $a,b,c$ be the residues of $\cot \pi z$ at these points (and these are fairly easy to obtain), then

$$\cot \pi z -\left (\frac{a}{z}+\frac{b}{z-1}+\frac{c}{z+1}\right )$$

is analytic on $\{|z|<2\}.$ Thus the expression in parentheses will be the desired principal part. So all we need to do is find the Laurent expansions of $1/z,1/(z-1),1/(z+1)$ in the annulus $\{1<|z|<2\}.$ This is straightforward.

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  • $\begingroup$ that's very nice. thanks $\endgroup$ – Ashley Jun 1 '15 at 18:21

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