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I hope to evaluate the contour integral of $\displaystyle\int_\gamma \frac{z}{\sin z}dz$

where $\gamma$ is circle of radius $\frac{3\pi}{2}$ centered at $z = 0$ and oriented clockwise.

I have trouble evaluating the residue at $z = 0$. Even if I can evaluate the residue I wonder if the answer is just $2\pi i$ times the residues. I also wonder why I have to evaluate with this fixed value of radius. Is the integral is zero even if clockwise?

Thanks in advance.

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    $\begingroup$ The residue at $0$ is $0$, because $z/\sin(z)$ has no singularity at $0$. You should worry about the residues at $\pi$ and $-\pi$. $\endgroup$ – J.R. Nov 22 '15 at 9:51
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    $\begingroup$ Notice that $\frac{z+\pi}{\sin(z+\pi)}=-\frac{z}{\sin z}-\frac{\pi}{z}\cdot\frac{z}{\sin z}$, hence it is pretty easy to compute the residues at $\pm \pi$. $\endgroup$ – Jack D'Aurizio Nov 22 '15 at 12:44
  • $\begingroup$ @Your Ad Here how do you know that and how to calculate the residue in this case? $\endgroup$ – Ka-Wa Yip Nov 23 '15 at 1:32
  • $\begingroup$ @Jack D'Aurizio i don't understand. Can I use L hospital rule? $\endgroup$ – Ka-Wa Yip Nov 23 '15 at 3:19
  • $\begingroup$ Is the integral is zero? $\endgroup$ – Ka-Wa Yip Nov 23 '15 at 3:20
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First you need to draw a good contour for this in order to locate all the corresponding poles. The poles inside your cyrcle are $z_1=0, z_2=\pi , z_3=-\pi$ . Recall that if $f(z)=\frac{g(z)}{h(z)}$ then the residue at any pole can be calculated by the following formula: $$ Res(f,z_0)=\frac{g(z_0)}{h^{'}(z_0)}$$ One shall use the formula stated above when the Laurent expansion of the integrand is difficult to find. So now we are ready to proceed: $$\int_{\gamma}\frac{z}{\sin(z)}dz=2 \pi i (Res(f,z_1)+Res(f,z_2)+Res(f,z_3))$$ $Res(f,z_1)=\frac{0}{\cos(0)}=0$, $Res(f,z_2)=\frac{\pi}{\cos(\pi)}=-\pi$, $Res(f,z_3)=\frac{-\pi}{\cos(-\pi)}=\pi$ Calculating the sum of residues gives the result which is $0$ .

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