Partial fraction expansion using residue theorem

Question

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?

what have you tried? how is the path defined? what's wrong with just doing a partial fraction decomposition without using the residue theorem? — thesmallprint, Nov 13 '17 at 18:06
@thesmallprint The question asks me to use residue theorem only. I think I should take a path containing the singularities $1, 2, 5$, but on any such path I do not know what the contour integral of $f$ will be; and that also will not give me '$f$' in a partial fraction form — ab123, Nov 13 '17 at 18:10
@ab123 https://en.wikipedia.org/wiki/Partial_fraction_decomposition#Residue_method — Math Lover, Nov 13 '17 at 18:13
@MathLover Ok so in this case it will be $1+ 9/(z-1) -28/(z-2) +35/(z-5)$ — ab123, Nov 13 '17 at 18:32
@ab123 I haven't done the calculations. You can verify whether your decomposition is correct by combining the fractions into one fraction. If that matches $f(z)$ then you are done. — Math Lover, Nov 13 '17 at 18:38

score 7 · Accepted Answer · answered Nov 14 '17 at 05:29

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 $2,5$.

Can you please explain why $r(z)$ has only removable singularities? — ab123, Nov 14 '17 at 08:16
@ab123 : Because $r(z)$ remains bounded nearly the possible finite singularities at $1,2,5$. — Disintegrating By Parts, Nov 14 '17 at 08:20

thesmallprint · Answer 2 · 2017-11-13T23:15:50.780

1

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 $$\gamma=2\pi i(9-28+35)=2\pi i (16)=32\pi i.$$

edited Nov 13 '17 at 23:15

answered Nov 13 '17 at 18:48

thesmallprint

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Partial fraction expansion using residue theorem

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