Partial fractions with complex linear terms

  • #1
zenterix
555
76
Homework Statement
Decompose

$$\frac{5s+6}{(s^2+4)(s-2)}$$
Relevant Equations
using partial fractions.
I am interested specifically in solving this problem by factoring the quadratic term into complex linear factors.

$$s^2+4=0$$

$$\implies s=\pm 2i$$

$$\frac{5s+6}{(s-2i)(s+2i)(s-2)}=\frac{A}{s-2i}+\frac{B}{s+2i}+\frac{C}{s-2}$$

We can solve for ##C## using the cover-up method with ##s=2## to find ##C+2##.

Thus

$$\frac{5s+6}{(s-2i)(s+2i)(s-2)}=\frac{A}{s-2i}+\frac{B}{s+2i}+\frac{2}{s-2}$$

From what I read, we can use the cover-up method here as well.

I tried to solve for ##A## by setting ##s=2i## as below

$$\left .\frac{5s+6}{(s+2i)(s-2)}\right |_{s=2i}=A$$

$$=\frac{10i+6}{4i(2i-2)}$$

$$=\ldots$$

$$=\frac{i-1}{-4}$$

The notes I am reading say

At the end, conjugate complex terms have to be combined in pairs to produce real summands, and the calculations can sometimes be longer.

I'm not sure how to proceed to get a real number for the coefficient ##A##.
 
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  • #2
You cannot get a real ##A##. That’s why you need to combine it with the conjugate term to get a real expression.
 
  • #3
In the context of this specific problem, what does that look like exactly?

After all, I am after ##A## and ##B## and I found ##A## to be complex (I also found the complex ##B=\frac{4-i}{-4}##). What does it mean to combine this complex ##A## with its conjugate in the context of the partial fractions decomposition?
 
  • #4
zenterix said:
Homework Statement: Decompose

$$\frac{5s+6}{(s^2+4)(s-2)}$$
Relevant Equations: using partial fractions.
If you're just interested in decomposing the fraction above, the usual way to do it is like so:
##\frac{5s+6}{(s^2+4)(s-2)} = \frac {As + B}{s^2 + 4} + \frac C {s - 2}##

As it turns out, all three coefficients are real. Note that when a denominator in the fraction you want to decompose is an irreducible quadratic (that is, irreducible over the reals), the corresponding fraction will have a numerator that is linear with two terms.

I wrote an Insights article about this subject about 8 years ago -- https://www.physicsforums.com/insights/partial-fractions-decomposition/.
 
  • #5
Mark44 said:
If you're just interested in decomposing the fraction above, the usual way to do it is like so:
Yes, I've done that. At this point I've done various exercises for each of the cases in your article. But I would like to know how to do it specifically by having complex linear factors in the denominator.

The notes where I encountered this possibility are here.
 
  • #6
zenterix said:
In the context of this specific problem, what does that look like exactly?
Like this:

Mark44 said:
##\frac{5s+6}{(s^2+4)(s-2)} = \frac {As + B}{s^2 + 4} + \frac C {s - 2}##

You will not get real coefficients with linear denominators.
 
  • #7
Alternatively you simply note that the imaginary parts between the complex denominators cancel when ##s## is real.
 
  • #8
zenterix said:
But I would like to know how to do it specifically by having complex linear factors in the denominator.
Your decomposition to linear factors is:$$\frac{\left(5s+6\right)}{\left(s^{2}+4\right)\left(s-2\right)}=-\frac{1+\frac{i}{4}}{s-2i}-\frac{1-\frac{i}{4}}{s+2i}+\frac{2}{s-2}$$The first and second terms on the right are complex conjugates, so as long as ##s\in\mathbb{R}## they add to a real result. What more do you want?
 
  • #9
@renormalize

I made a mistake in computing ##B## and didn't see that ##A## and ##B## turn out to be complex conjugates.
 
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