"आंशिक भिन्न" (अनुभाग) सम्पादित करें

===उदाहरण ४ (residue method)===

:<math> f(z)=\frac{z^{2}-5}{(z^2-1)(z^2+1)}=\frac{z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}</math>

Thus, ''f''(''z'') can be decomposed into rational functions whose denominators are ''z''+1, ''z''−1, ''z''+i, ''z''−i. Since each term is of power one, −1, 1, −''i'' and ''i'' are simple poles.

Hence, the residues associated with each pole, given by
:<math>\frac{P(z_i)}{Q'(z_i)} = \frac{z_i^2 - 5}{4z_i^3}</math>,
are 
:<math> 1, -1, \tfrac{3i}{2}, -\tfrac{3i}{2}</math>,
respectively, and

:<math> f(z)=\frac{1}{z+1}-\frac{1}{z-1}+\frac{3i}{2}\frac{1}{z+i}-\frac{3i}{2}\frac{1}{z-i}</math>.