I have to use residue theorem to find the partial fraction expansion of
$$f(z) = \frac{(z+1)(z+2)(z+5)}{(z-1)(z-2)(z-5)}$$
I can see that the function has singularities at $z=1, 2, 5$
I know what the residue theorem is, but it contains a contour integration of $f$ on a closed path containing some singularities. How shall I get the partial fractions from it?
$\endgroup$ 52 Answers
$\begingroup$You know that there are order one singularities of $f$ at $z=1,2,5$. Therefore, with the correct constants $A,B,C$, the function $$ r(z)=f(z) - \frac{A}{z-1}-\frac{B}{z-2}-\frac{C}{z-5} $$ has only removable singularities in the finite plane. And, because $r$ is bounded near $\infty$, it follows that $r$ extends to a bounded entire function of $z$, which makes it a constant. That constant is $\lim_{z\rightarrow\infty}f(z)=1$. So, $$ f(z) = 1+\frac{A}{z-1}+\frac{B}{z-2}+\frac{C}{z-5}. $$ The constant $A$ is determined by a residue at $z=1$: $$ A=\lim_{z\rightarrow 1}(z-1)f(z) = \frac{2\cdot 3\cdot 6}{(-1)(-4)}=9. $$ The constants $B$ and $C$ are similarly determined by residues at $2,5$.
$\endgroup$ 2 $\begingroup$Your partial fraction decomposition is correct. So our residues are: $9, -28, 35.$
Then, by the residue theorem we get that our integral, call it $\gamma$ is $$\gamma=2\pi i(9-28+35)=2\pi i (16)=32\pi i.$$
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