Show that $ \Gamma(1-z)=-\frac{1}{z-1}+\cdots $

Question

I am a bit confusing about this notation. How to show that ? $$ \Gamma(1-z)=-\frac{1}{z-1}+\cdots $$ Reference:

http://www.math.harvard.edu/archive/213b_spring_05/functional_equation_by_residues.pdf

Answer 1

The intent was most likely that they were showing the terms of the Laurent expansion at negative powers to obtain the residues. Note that:

\begin{align}\Gamma(1-z)&=\frac{\Gamma(2-z)}{1-z}\\&=\frac{\Gamma(1)-\Gamma'(1)(z-1)+\mathcal O(z-1)^2}{1-z}\\&=\frac1{1-z}+\Gamma'(1)+\mathcal O(z-1)\end{align}

as $z\to1$.