Editing Residue (complex analysis) (section)

==== Example 2 ====
As a second example, consider calculating the residues at the singularities of the function<math display="block">f(z) = {\sin z \over z^2-z}</math>which may be used to calculate certain contour integrals. This function appears to have a singularity at ''z'' = 0, but if one factorizes the denominator and thus writes the function as<math display="block">f(z) = {\sin z \over z(z - 1)}</math>it is apparent that the singularity at ''z'' = 0 is a [[removable singularity]] and then the residue at ''z'' = 0 is therefore 0.
The only other singularity is at ''z'' = 1. Recall the expression for the Taylor series for a function ''g''(''z'') about ''z'' = ''a'':<math display="block"> g(z) = g(a) + g'(a)(z-a) + {g''(a)(z-a)^2 \over 2!} + {g'''(a)(z-a)^3 \over 3!}+ \cdots</math>So, for ''g''(''z'') = sin&nbsp;''z'' and ''a'' = 1 we have<math display="block"> \sin z = \sin 1 + (\cos 1)(z-1)+{-(\sin 1)(z-1)^2 \over 2!} + {-(\cos 1)(z-1)^3 \over 3!} + \cdots.</math>and for ''g''(''z'') = 1/''z'' and ''a'' = 1 we have<math display="block"> \frac{1}{z} = \frac1 {(z - 1) + 1} = 1 - (z - 1) + (z - 1)^2 - (z - 1)^3 + \cdots.</math>Multiplying those two series and introducing 1/(''z''&nbsp;−&nbsp;1) gives us<math display="block"> \frac{\sin z} {z(z - 1)} = {\sin 1 \over z-1} + (\cos 1 - \sin 1) + (z-1) \left(-\frac{\sin 1}{2!} - \cos1 + \sin 1\right) + \cdots.</math>So the residue of ''f''(''z'') at ''z'' = 1 is sin&nbsp;1.