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The function I have is $f(z)=\zeta(z)\Gamma(z − 1)\sin(\pi z)$ and I need to find all singularities and their residues so I can evaluate a clockwise contour integral for the contour $\left\lvert z+\frac{1}{2}\right\rvert=4.$

So far I've simplified $f(z)$ using the identity $\Gamma(z)=(z-1)\Gamma(z − 1)$ and then with Euler's reflection formula $\Gamma(z)\sin(\pi z)=\frac{\pi}{\Gamma(1-z)}$ to get $f(z)=\zeta(z)\pi /(\Gamma(1-z))(z-1)).$

$\zeta(z)$ has a pole at $z=1$, so combined with the $\frac{1}{z-1}$, $f(z)$ has a double pole at $z=1$. Is my logic correct so far? I'm not sure how to calculate the residue now. I know $\zeta(z)$ has a residue of $1$ at $z=1,$ unsure where to go from here.

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  • $\begingroup$ $\Gamma(1-z)$ also has a pole at $z = 1$, so $\frac{1}{\Gamma(1-z)}$ has a (simple) zero there. That cancels one order. It's IMO easier to look at the factor $\Gamma(z-1)\sin (\pi z)$. You know where $\Gamma$ has poles, and you know how $\sin (\pi z)$ behaves there. That shows that $\Gamma(z-1)\sin (\pi z)$ is rather well-behaved. $\endgroup$ – Daniel Fischer May 10 '16 at 15:13
  • $\begingroup$ Oh. So the sine function cancels out all the poles of the Gamma function, leaving only the simple pole from the Riemann-zeta function? $\endgroup$ – John May 10 '16 at 15:21
  • $\begingroup$ Aye. Then you just need to find the residue there. $\endgroup$ – Daniel Fischer May 10 '16 at 15:24
  • $\begingroup$ So then the residue is just 1? I don't need to take into account the Gamma and sine functions? $\endgroup$ – John May 10 '16 at 15:33
  • $\begingroup$ No, I meant you need to find the residue of $\zeta(z) \Gamma(z-1)\sin (\pi z)$ at $1$. It's not equal to the residue of $\zeta$ there. $\endgroup$ – Daniel Fischer May 10 '16 at 15:35

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