Question #235597

find the residue of f(z) =1/(z2+1)2 at z=i


1
Expert's answer
2021-09-14T06:06:10-0400

f(z)=1(z2+1)2=1(z+i)2(zi)2f(z) = \frac{1}{{{{\left( {{z^2} + 1} \right)}^2}}} = \frac{1}{{{{\left( {z + i} \right)}^2}{{\left( {z - i} \right)}^2}}}

z = i is the zero of the order 2 denominator, so z = i is the second order pole.

Then

resz=if(z)=limzi(1(z+i)2(zi)2(zi)2)=limzi(1(z+i)2)=limzi2(z+i)3=2(2i)3=28i=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}

Answer: resz=if(z)=i4\mathop {res}\limits_{z = i} f(z) = - \frac{i}{4}


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