# Algebraic Expansion with complex numbers

h_ngm_n
Hi all,

I am a bit rusty and have hit a snag with decomposition of partial fractions, I am taking an Engineering course dealing with Laplace transforms. The example is:

F(s)= 3 / s(s2+2s+5)

Now I get that there are complex roots in the denominator and that there are conjugate complex roots (s+1±2i) giving:

F(s)= 3 / s(s+1-2i)(s+1-2i)

So partial fraction decomposition would give:

F(s) = K1/s + K2/(s+1+2i) + K3/(s+1-2i)

Ok, so solving for K1= 3/5 is easy and I get that, but when it comes to K2 which involves substituting (-1-2i) in for 's' and then expanding, I can't seem to get the answer. What I did was:

K2=3[STRIKE](s+1+2i)[/STRIKE]/s(s+1-2i)[STRIKE](s+1+2i)[/STRIKE]
K2=3/s(s+1-2i)
K2=3/(-1-2i)[(-1-2i)+1-2i)]

It is the algebraic expansion on the denominator that is getting me. When I do it I get:

(-1-2i)[(-1-2i)+1-2i)]
= (-1-2i)(0-4i)
=-1(-4i)-2i(-4i)
=4i+8i2
=8+4i
=4(2+i)

However my textbook gets -3/20 (2+i) for the final partial fraction... I'm lost any help would be greatly appreciated.

Homework Helper
Gold Member
4i+8i2 = 4i-8 = [-8+4i, edit]

Last edited:
h_ngm_n
Ah yes, simple error... my bad. But it still doesn't match the answer in my text :(

Homework Helper
Ah yes, simple error... my bad. But it still doesn't match the answer in my text :(
That's because you haven't simplified K2 completely. With lewando's correction the denominator simplifies to -8+4i, so K2 is:
$K_2 = \frac{3}{-8 + 4i}$
Multiply the top and bottom by the conjugate of the denominator, and eventually, you will get:
$K_2 = -\frac{3}{20}\left( 2 + i \right)$

h_ngm_n
Ok, thank you for your time, I needed to simplify my fraction to get the complex term out of the denominator didn't I?

Homework Helper
Ok, thank you for your time, I needed to simplify my fraction to get the complex term out of the denominator didn't I?

Yes, the fraction wouldn't be considered simplified otherwise.

Millennial
It seems simpler to me to directly find the residues and then sum them to obtain the inverse Laplace, given that all poles here are simple. I do not know if you have any knowledge of the inverse formula, but it is certainly easier to apply here.

h_ngm_n
Hi Millennial,

So say that the residues are (took me a while to get this :uhh::)

K1=-0.4615
K2=0.2308-j2.0110
K3=0.2308+j2.0110

I can make this

g(t)= -0.4615e(-1.4t)+[(0.2308-j2.0110) e(-(1-j0.7)t)+(0.2308+j2.0110) e(-(1+j0.7)t) ] just by 'inspecting' the residues ? As opposed to breaking the partial fractions down and then doing inverse Laplace transform?

And I am little confused with the method to express it as trig identities:

The formulae: cosθ=(e+e-jθ)/2

and: sinθ=(e-e-jθ)/2j

I get a bit stuck using this to convert from exponential form to trig form...

Of course, you can avoid all the complex number stuff by using the partial fraction expansion$$\frac 3 {s(s^2+2s+5)}= \frac A s + \frac{Bs+C}{s^2+2s+5}$$and writing ##s^2+2s+5=(s+1)^2+4## when taking the inverse Laplace Transform.