Given $$f(z)=\dfrac{\tanh (2z)}{z\sin(\pi z)}$$ Are $z=\pm 1$ simple poles of $f(z)$? It appears so as $\sin(n\pi)=0$ for integer n. But the integral along the circle $|z|=\pi/2$ of this function is given as $4i$. Did I go wrong?
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Yes, they are poles of order $1$ but the residues at these points cancel out. It has a simple pole at $0$. The residue is $\frac 2 {\pi}$ so the integral is $2\pi i \frac 2 {\pi}=4i$.
Kavi Rama Murthy
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Is it a simple pole at zero? – Purushothaman Aug 01 '19 at 09:54
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No. it is a pole of order $3$. – Kavi Rama Murthy Aug 01 '19 at 09:55
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@Kavi Rama Murthy Sir..the numerator also has a zero at $z=0$. – Nitin Uniyal Aug 01 '19 at 11:40
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$\tanh (2z) =\dfrac{e^z-e^{-z})}{e^z+e^-{z}}$ – Purushothaman Aug 02 '19 at 00:08
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@Purushothaman You are right. There is a simple pole at $0$. The residue is $\frac 2 {\pi}$ so the integral is $2\pi i \frac 2 {\pi}=4i$. – Kavi Rama Murthy Aug 02 '19 at 00:14
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$$f(z)=\dfrac{\tanh (2z)}{z\sin(\pi z)}=g(z)h(z)$$ where $g(z)=\frac{\tanh (2z)}{z}$ and $h(z)=\frac{1}{sin(\pi z)}$.
Note that $g(z)$ had removable singularity at $z=0$ while $h(z)$ has simple pole there, so $f(z)$ also has simple pole at $z=0$ in addition to the simple poles at $z=\pm 1$
Nitin Uniyal
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