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Let's suppose that we're given a holomorphic function

$$f(z) = \dfrac{1}{\sin(z)}$$

And we want to find the coefficient of 20th term in Laurent series. I was trying to apply that $$a_k = \dfrac{1}{2\pi i}\int_{\Gamma} \dfrac{f(z)}{(z-a)^{k+1}}dz$$

for $k = 20$,

$$a_{20} = \dfrac{1}{2\pi i}\int_{\Gamma} \dfrac{f(z)}{z^{21}}dz = \dfrac{1}{2\pi i}\int_{\Gamma} \dfrac{1}{\sin(z)z^{21}}dz$$

Is there a way to evaluate this contour integral? And can we use residue theorem?

Melz
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    I think you need to apply $$Res(f;z_0) = \lim_{z \to z_0}\frac{1}{(m-1)!} \frac{d^{m-1}}{dz^{m-1}} [(z-z_0)^m f(z)] $$ – IAmNoOne Nov 09 '19 at 22:20
  • What is $\Gamma$--- a small circle around $0$? The $\sin z$ term in the denominator is going to be hard to integrate; it's probably easier to just invert $\sin z/z= 1 - \frac{1}{6} z^2 + \dots \in \mathbb{C}[[z]]$. – anomaly Nov 09 '19 at 22:20
  • Can we use residue theorem? – Melz Nov 09 '19 at 22:23
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    $f(z)$ is an odd function. In its Laurent expansion, all coefficients for even powers are zero. – achille hui Nov 09 '19 at 22:37
  • You already used the residue theorem to argue that $a_k$ was equal to that integral. – Aaron Nov 10 '19 at 04:02

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Use the formula $g^{(n)} (0)=\frac {n!} {2\pi i} \int_{\Gamma} \frac {g(z)} {z^{n+1}} dz$. Take $n=21$ and $g(z)=\frac z {\sin z}$ which is analytic in side the unit circle.

Kavi Rama Murthy
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  • Can we use the fact that $Res(f;z_0) = \lim_{z \to z_0}\frac{1}{(m-1)!} \frac{d^{m-1}}{dz^{m-1}} [(z-z_0)^m f(z)]$? – Melz Nov 10 '19 at 08:07
  • @Melz Yes, you can and you get the same answer whether you use Residue Theorem or Cauchy's integral formula for the derivatives. – Kavi Rama Murthy Nov 10 '19 at 08:12