If f(z) has a pole of order k at z=z0 then
rez(f,z0)=z→z0lim((k−1)!1dzk−1dk−1((z−z0)kf(z)))
ez=1+1!z+2!z2+3!z3+...
ez−1=1!z+2!z2+3!z3+...
z→0lim(zf(z))=z→0lim(z(z(ez−1)1))=z→0lim(ez−11)
=does not exist
(2−1)!1dz2−1d2−1((z−0)2f(z))
=dzd(z2(z(ez−1)1))
=dzd(ez−1z)
=(ez−1)2ez−1−zez
rez(f,0)=z→0lim((ez−1)2ez−1−zez)
=z→0lim(2ez(ez−1)ez−ez−zez)
=z→0lim(2(ez−1)−z)
=z→0lim(2ez−1)=−21
rez(f,0)=−21
Comments
Thank you so much for your answer.