Partial-fraction expansions of cot and 1/sin

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(a)

Prove that

π2sin⁡2πz=∑n=−∞∞1(z−n)2. \frac{\pi ^2}{\sin ^2\pi z} = \sum _{n=-\infty }^{\infty } \frac{1}{(z-n)^2}.
(b)

Use this to show that

πcot⁡πz=1z+∑n=1∞2zz2−n2. \pi \cot \pi z = \frac1z + \sum _{n=1}^{\infty } \frac{2z}{z^2-n^2}.

Justify your steps.

(c)

Develop πcot⁡πz\pi \cot \pi z in a Laurent series about the origin directly and by use of (b), with enough terms to find the values of ∑n=1∞1n2\sum_{n=1}^{\infty } \frac{1}{n^2} and ∑n=1∞1n4\sum_{n=1}^{\infty } \frac{1}{n^4}.

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

1sin⁡z=1z+∑n=1∞(−1)n2zz2−n2π2. \frac{1}{\sin z} = \frac{1}{z}+\sum _{n=1}^{\infty } \frac{(-1)^{n}2z}{z^{2}-n^{2}\pi ^{2}}.
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Expand the function f(x)f(x) into a Fourier series on the given interval, the length of the interval being the period. f(x)=cos⁡ax,a∉Zf(x) = \cos a x, a \notin Z on the segment [−π;π][-\pi ; \pi ]. Using the resulting expansion, prove that

ctg⁡x=1x+∑n=1∞2xx2−πn2 \operatorname {ctg} x = \frac{1}{x}+\sum _{n = 1}^{\infty } \frac{2 x}{x^{2}-\pi n^{2}}
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Prove that for any x∈R,x≠πm,m∈Zx \in R, x \neq \pi m, m \in Z, the equality holds

1sin⁡x=1x∑n=1∞(−1)n(1x−πn+1x+πn) \frac{1}{\sin x} = \frac{1}{x} \sum _{n = 1}^{\infty }(-1)^{n}\left(\frac{1}{x-\pi n}+\frac{1}{x+\pi n}\right)
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References

[CJP+11] Chen J., Jiang G., Pan Y., Qin T., Tong Y., Wu Q., Xu S., Li T. Problems and Solutions in Mathematics. 2nd ed., World Scientific, 2011.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Интегралы. Ряды. 2-е изд., ФИЗМАТЛИТ, 2003.