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I know that the $\Gamma(z)$ has simple poles at $0, -1, -2,...$

Does that mean $\Gamma(z/2)$ has simples pole at $0, -1/2,-1,...$?

Also, $Res(\Gamma,-k)=(-1)^k/k!$, so is the residue of $\Gamma(z/2)$ the same but with $k/2$?

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    $\begingroup$ I'd say you'd have to multiply the original poles by two to get the poles of $\;\Gamma\left(\frac z2\right)\;$ , not divide them by two... $\endgroup$ – DonAntonio Mar 21 '18 at 20:10
  • $\begingroup$ Sorry, can you explain why? How does that affect the residue function? $\endgroup$ – Sam Mar 21 '18 at 20:16
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If $\;k\;$ is a pole of $\;\Gamma(z)\;$ , then

$$0=\frac1{\Gamma(k)}=\frac1{\Gamma\left(\frac{2k}2\right)}\implies 2k\;\text{ is a pole of}\;\;\Gamma\left(\frac z2\right)$$

Now you try to work out the residues...

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