Editing Cauchy's integral formula (section)

== Example ==
[[File:ComplexResiduesExample.png|thumb|300px|Surface of the real part of the function {{math|1=''g''(''z'') = {{sfrac|''z''<sup>2</sup>|''z''<sup>2</sup> + 2''z'' + 2}}}} and its singularities, with the contours described in the text.]]
Let
<math display="block">g(z) = \frac{z^2}{z^2+2z+2},</math>
and let {{math|''C''}} be the contour described by {{math|1={{abs|''z''}} = 2}} (the circle of radius 2).

To find the integral of {{math|''g''(''z'')}} around the contour {{math|''C''}}, we need to know the singularities of {{math|''g''(''z'')}}. Observe that we can rewrite {{math|''g''}} as follows:
<math display="block">g(z) = \frac{z^2}{(z-z_1)(z-z_2)}</math>
where {{math|1=''z''<sub>1</sub> = − 1 + ''i''}} and {{math|1=''z''<sub>2</sub> = − 1 − ''i''}}.

Thus, {{math|''g''}} has poles at {{math|''z''<sub>1</sub>}} and {{math|''z''<sub>2</sub>}}. The [[absolute value|moduli]] of these points are less than 2 and thus lie inside the contour. This integral can be split into two smaller integrals by [[Cauchy–Goursat theorem]]; that is, we can express the integral around the contour as the sum of the integral around {{math|''z''<sub>1</sub>}} and {{math|''z''<sub>2</sub>}} where the contour is a small circle around each pole<!-- diagram works best -->. Call these contours {{math|''C''<sub>1</sub>}} around {{math|''z''<sub>1</sub>}} and {{math|''C''<sub>2</sub>}} around {{math|''z''<sub>2</sub>}}.

Now, each of these smaller integrals can be evaluated by the Cauchy integral formula, but they first must be rewritten to apply the theorem. For the integral around {{math|''C''<sub>1</sub>}}, define {{math|''f''<sub>1</sub>}} as {{math|1=''f''<sub>1</sub>(''z'') = (''z'' − ''z''<sub>1</sub>)''g''(''z'')}}. This is [[holomorphic function|analytic]] (since the contour does not contain the other singularity). We can simplify {{math|''f''<sub>1</sub>}} to be:
<math display="block">f_1(z) = \frac{z^2}{z-z_2}</math>
and now
<math display="block">g(z) = \frac{f_1(z)}{z-z_1}.</math>

Since the Cauchy integral formula says that:
<math display="block">\oint_C \frac{f_1(z)}{z-a}\, dz=2\pi i\cdot f_1(a),</math>
we can evaluate the integral as follows:
<math display="block">
  \oint_{C_1} g(z)\,dz
 =\oint_{C_1} \frac{f_1(z)}{z-z_1}\,dz
 =2\pi i\frac{z_1^2}{z_1-z_2}.
</math>

Doing likewise for the other contour:
<math display="block">f_2(z) = \frac{z^2}{z-z_1},</math>
we evaluate
<math display="block">
  \oint_{C_2} g(z)\,dz
 =\oint_{C_2} \frac{f_2(z)}{z-z_2}\,dz
 =2\pi i\frac{z_2^2}{z_2-z_1}.
</math>

The integral around the original contour {{math|''C''}} then is the sum of these two integrals:
<math display="block">\begin{align}
     \oint_C g(z)\,dz
&{}= \oint_{C_1} g(z)\,dz
   + \oint_{C_2} g(z)\,dz \\[.5em]
&{}= 2\pi i\left(\frac{z_1^2}{z_1-z_2}+\frac{z_2^2}{z_2-z_1}\right) \\[.5em]
&{}= 2\pi i(-2) \\[.3em]
&{}=-4\pi i.
\end{align}</math>

An elementary trick using [[partial fraction decomposition]]:
<math display="block">
  \oint_C g(z)\,dz
 =\oint_C \left(1-\frac{1}{z-z_1}-\frac{1}{z-z_2}\right) \, dz
 =0-2\pi i-2\pi i
 =-4\pi i
</math>