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$\int_C \frac{1}{\sin z}$ where $C := \{z \in C : |z| = 1 \}$ I know $\frac{1}{\sin z}$ has a pole at $z = 0$. I can apply Cauchy Residue Theorem, so I need to know the laurent series expansion of $\frac{1}{\sin z}$ which I am unable to figure out.

Is there any other neat way to approach this?

Thanks in advance.

bob
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Dwaipayan Sharma
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    https://math.stackexchange.com/questions/648923/calculate-laurent-series-for-1-sinz – Volk Mar 19 '24 at 19:11

3 Answers3

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Since you just want the residue, you can write $\sin z=zg(z)$ for some entire $g$, where $g(0)\ne0$. Then by residue theorem, your integral is $$\dfrac{2\pi\textbf{i}}{g(0)}=2\pi\textbf{i}\displaystyle\lim_{z\to0}\frac{1}{g(z)}=2\pi\textbf{i}\displaystyle\lim_{z\to0}\dfrac{z}{\sin z}$$ So you can get the residue without computing the Laurent series of $1/\sin z$.

Angae MT
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Let $$f(z)=\frac{z}{\sin z},0<|z|<2,\quad f(0)=1,$$ then Morera's theorem implies $f$ is analytic in $\{z\mid |z|<2\}$. So, by Cauchy integral formula, we have $$\int_{|z|=1}\frac{dz}{\sin z}=\int_{|z|=1}\frac{f(z)}{z}dz =2\pi if(0)=2\pi i.$$

Riemann
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Hint: $\dfrac{1}{\sin z} - \dfrac{1}{z}\;$ is analytic inside $C$, so $\int_C \left(\dfrac{1}{\sin z} - \dfrac{1}{z}\right)dz=0$.

To see that $\dfrac{1}{\sin z} - \dfrac{1}{z}=\dfrac{z-\sin z}{z\sin z}\;$ is analytic, note that the leading term in the numerator is $\frac16x^3$ and the leading term in the denominator is $x^2$.

TonyK
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