Editing Contour integration (section)

====Using the method of residues====
Consider the [[Laurent series]] of {{math|''f''(''z'')}} about {{mvar|i}}, the only singularity we need to consider. We then have
<math display=block>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</math>

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

It is clear by inspection that the residue is {{math|−{{sfrac|''i''|4}}}}, so, by the [[residue theorem]], we have
<math display=block>\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</math>

Thus we get the same result as before.