Residues of $\Gamma$ using Euler's reflection formula

Question

We know Euler's reflection formula to be:

$\Gamma (z)\Gamma (1-z)=\frac{\pi}{\sin (\pi z)}$

I know how to obtain the residues for $\Gamma$ in $\{0,-1,-2,...\}$ using the equation $\Gamma (z+1)=z\Gamma (z)$ but,

is there a way to calculate the residues using Euler's formula?

Answer 1

Let's calculate the residue of $\frac{\pi}{\sin{\pi z}}$ at $z=-n$, for $n\in \mathbb{Z}_{\ge0}$. Proceeding we have

$$\begin{align} \text{Res}\left(\frac{\pi}{\sin{\pi z}}, z=-n\right)&=\lim_{z\to -n}\frac{\pi (z+n)}{\sin{\pi z}}\\\\ &=(-1)^n\tag 1 \end{align}$$

Next, note that since $\Gamma(z)$ has no singularity on the positive real axis, then

$$\begin{align} \text{Res}\left(\Gamma(z)\Gamma(1-z), z=-n\right)&=\Gamma(1+n)\text{Res}\left(\Gamma(z), z=-n\right)\\\\ &=n! \text{Res}\left(\Gamma(z), z=-n\right)\tag2 \end{align}$$

Putting $(1)$ and $(2)$ together yields the coveted answer

$$\text{Res}\left(\Gamma(z), z=-n\right)=\frac{(-1)^n}{n!}$$