find the residue of f(z) =1/(z2+1)2 at z=i
"f(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
"\\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: "\\mathop {res}\\limits_{z = i} f(z) = - \\frac{i}{4}"
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