Wolfram Alpha gives me a pole at $z=0$ of order $3$. Why is this the case? The denominator is a polynomial of order $2$?
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7The numerator has also a pole at the origin ... – Martin R Feb 12 '23 at 11:07
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This MSE question may help you. – Carlos Adir Feb 12 '23 at 11:19
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2$\frac{\pi\cot(\pi z)}{z^2} = \frac{\pi \cos(\pi z)}{z^2\sin(\pi z)}$. Now you can see it clearly. – CroCo Feb 12 '23 at 11:39
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Hint: are you aware $\lim_{z\to0}\pi z\cot\pi z=1$? – J.G. Feb 14 '23 at 16:45
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HINT: Find the least $n\in\mathbb N$ such that $z^nf(z)$ is finite as $z\rightarrow 0$.
Nitin Uniyal
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It is $\cot(x) = \sum\limits_{n=1}^{\infty}\frac{2^{2n+1}}{\pi^{2n}}\cdot \lambda(2n)\cdot x^{2n-1}$. So use $x = \pi z$ and we have $$ \frac{\pi\cdot\cot(\pi z)}{z^2} = \pi \cdot \frac{1}{z^2}\sum\limits_{n=1}^{\infty} \frac{2^{2n+1}}{\pi^{2n}}\lambda(2n)\pi^{2n-1}z^{2n-1} $$ and shortening leads to $$ \frac{\pi\cdot\cot(\pi z)}{z^2} = \sum\limits_{n=1}^{\infty}2^{2n+1}\lambda(2n)z^{2n-3} = \frac{1}{z^3}\sum\limits_{n=1}^{\infty}2^{2n+1}\lambda(2n)z^{2n} $$ So for $k\in\{1,2\}$ the series $z^k\cdot \frac{\pi\cdot\cot(\pi z)}{z^2} $ is not defined for $z=0$. But $k=3$ does the trick.
mathquester
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