The poles of f(z)=(z+1)2(z+4)z2+2z are obtained from (z+1)2(z+4)=0.
We get, z=−1 is a pole of order 2 and z=−4 is a simple pole.
Res{f(z),−1}Res{f(z),−4}=z=−1limdzd{(z+1)2 f(z)}=z=−1limdzd{(z+1)2(z+1)2(z+4)(z2+2z)}=z=−1limdzd{z+4z2+2z}=z=−1lim{(z+4)2z2+8z+8}=91=z=−4lim{(z+4) f(z)}=z=−4lim{(z+4)(z+1)2(z+4)(z2+2z)}=z=−4lim{(z+1)2z2+2z}=0
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