information:extraction
Using the method of residues

Consider the Laurent series of f (z) about i, the only singularity we need to consider. We then have f (z) = − 1 4 (z − i) 2 + − i 4 (z − i) + 3 16 + i 8 (z − i) + − 5 64 (z − i) 2 + ⋯ {\displaystyle f(z)={\frac {-1}{4(z-i)^{2}}}+{\frac {-i}{4(z-i)}}+{\frac {3}{16}}+{\frac {i}{8}}(z-i)+{\frac {-5}{64}}(z-i)^{2}+\cdots }

(See the sample Laurent calculation from Laurent series for the derivation of this series.)

It is clear by inspection that the residue is − i / 4, so, by the residue theorem, we have ∮ C f (z) d z = ∮ C 1 (z 2 + 1) 2 d z = 2 π i Res z = i f (z) = 2 π i (− i 4) = π 2 ◻ {\displaystyle \oint _{C}f(z)\,dz=\oint _{C}{\frac {1}{\left(z^{2}+1\right)^{2}}}\,dz=2\pi i\,\operatorname {Res} _{z=i}f(z)=2\pi i\left(-{\frac {i}{4}}\right)={\frac {\pi }{2}}\quad \square }

Thus we get the same result as before.