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What is the path integral of $\frac{1}{zsin(z)}$ around $D_1(0)$? I've been trying to exploit residuals or use parametrization, but I can't come out of it.

What I actually need to do is to find the Laurent series of the function.

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If you meant $\,D_1(0)=S^1=\{z\in\Bbb C\;;\;|z|=1\}\,$ , then we have one unique singularity within this path, in $\,z=0\,$:

$$f(z):=\frac{1}{z\sin z}=\frac{1}{z}\cdot\csc z=\frac{1}{z}\left(\frac{1}{z}+\frac{z}{6}+\frac{7}{360}z^3+\ldots\right)=\frac{1}{z^2}+\frac{1}{6}+\frac{7}{360}z+\ldots$$ and thus

$$Res_{z=0}(f)=0\Longrightarrow\oint\limits_{S^1}f(z)\,dz=0$$

Added on request: Taylor series for $\,z\,$ "close" to zero:

$$\csc z=\frac{1}{\sin z}=\frac{1}{\left(z-\frac{z^3}{6}+\frac{z^5}{120}-\ldots\right)}=\frac{1}{z}\frac{1}{\left(1-\left(\frac{z^2}{6}-\mathcal O(z^4)\right)\right)}=$$

$$=\frac{1}{z}\left(1+\frac{z^2}{6}+\mathcal O(z^4)\right)=\frac{1}{z}+\frac{z}{6}+\mathcal O(z^3)$$

We stop there since we're interested only in the coefficient of $\,z^{-1}\,$ ...Note that we're relying on

$$\frac{1}{1-z}=1+z+z^2+\ldots\,\,\,,\,\,\text{for}\;\;|z|<1\;\;\text{(this is where the "close to zero" part kicks in)}$$

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Yes I meant that, I didn't knew the expression of csc(z). Can you show me how to, eventually "manually" find the first non zero terms of the Laurent series of f(z)? That was actually my issue. –  Temitope.A Feb 18 '13 at 23:09
    
I added some stuff to my answer, @Temitope.A –  DonAntonio Feb 18 '13 at 23:17
    
Am I right to think that finding the coefficient 7/360 with this method is difficult? –  Temitope.A Feb 18 '13 at 23:26
    
I've also tried writing $c_k=1/2\pi i \int_{S^1} f(z)(z-a)^{(-k-1)}$ but then get stuck calculating the path integral with parametrization or using the residual theorem (i.e for a pole of order 1 $c_{-1}=\lim _{z\rightarrow a}(z-a)f(z)$) –  Temitope.A Feb 18 '13 at 23:29
    
Not really: only take more summands in the "basic" step: $$\frac{1}{z}\frac{1}{1-\left(\frac{z^2}{6}-\frac{z^4}{120}\right)+\mathcal O(z^6)}=$$ $$1+\frac{z^2}{6}-\frac{z^4}{120}+\left(\frac{z^2}{6}-\frac{z^4}{120}\right)^2+ \ldots=$$ $$1+\frac{z^2}{6}-\frac{z^4}{120}+\frac{z^4}{36}-\ldots=1+\frac{z^2}{6}+ \frac{7}{360}z^4 +\ldots$$ Of course, all the above times $\,z^{-1}\,$... –  DonAntonio Feb 18 '13 at 23:35
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