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 $k/2$?
2026-03-25 01:21:18.1774401678
complex analysis- simple poles of the gamma function
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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...