- #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

I'm not sure how to proceed to get a real number for the coefficient ##A##.

$$s^2+4=0$$

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

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

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##.