The residue at z=1 of the function 2z/(1-z)(z+1)
The residue at z= 1 is
limz→ 1((z−1)2z(1−z)(z+1))=−2zz+1=limz→ 1(−2zz+1)Plug in the value z=1=−2⋅ 11+1=−1\lim _{z\to \:1}\left(\left(z-1\right)\frac{2z}{\left(1-z\right)\left(z+1\right)}\right)\\ =-\frac{2z}{z+1}\\ =\lim _{z\to \:1}\left(-\frac{2z}{z+1}\right)\\ \mathrm{Plug\:in\:the\:value}\:z=1\\ =-\frac{2\cdot \:1}{1+1}\\ =-1limz→1((z−1)(1−z)(z+1)2z)=−z+12z=limz→1(−z+12z)Pluginthevaluez=1=−1+12⋅1=−1
