find the residue of f(z) =1/(z2+1)2 at z=i
f(z)=1(z2+1)2=1(z+i)2(z−i)2f(z) = \frac{1}{{{{\left( {{z^2} + 1} \right)}^2}}} = \frac{1}{{{{\left( {z + i} \right)}^2}{{\left( {z - i} \right)}^2}}}f(z)=(z2+1)21=(z+i)2(z−i)21
z = i is the zero of the order 2 denominator, so z = i is the second order pole.
Then
resz=if(z)=limz→i(1(z+i)2(z−i)2⋅(z−i)2)′=limz→i(1(z+i)2)′=limz→i−2(z+i)3=−2(2i)3=−2−8i=14i=−i4\mathop {res}\limits_{z = i} f(z) = \mathop {\lim }\limits_{z \to i} {\left( {\frac{1}{{{{\left( {z + i} \right)}^2}{{\left( {z - i} \right)}^2}}} \cdot {{\left( {z - i} \right)}^2}} \right)^\prime } = \mathop {\lim }\limits_{z \to i} {\left( {\frac{1}{{{{\left( {z + i} \right)}^2}}}} \right)^\prime } = \mathop {\lim }\limits_{z \to i} \frac{{ - 2}}{{{{(z + i)}^3}}} = \frac{{ - 2}}{{{{(2i)}^3}}} = \frac{{ - 2}}{{ - 8i}} = \frac{1}{{4i}} = - \frac{i}{4}z=iresf(z)=z→ilim((z+i)2(z−i)21⋅(z−i)2)′=z→ilim((z+i)21)′=z→ilim(z+i)3−2=(2i)3−2=−8i−2=4i1=−4i
Answer: resz=if(z)=−i4\mathop {res}\limits_{z = i} f(z) = - \frac{i}{4}z=iresf(z)=−4i
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