How to Decompose (z^3+1)/(z(1-z)^2)?

[bryansteele]

Yes, another of these. How do you decompose (z³+1)/(z(1-z)²) ?

I've tried A/z + B/(1-z) + (Cz+D)/(1-z)² and a slew of others that don't work.

Thanks a bunch!

 

[BloodyFrozen]

(z³+1)/z(1-z²)

Do you mean:

(z³+1)/(z(1-z²))

or

((z³+1)/z)(1-z²)

 

[Dickfore]

First of all, the power of the numerator is the same as the power of the denominator. You have to make long division of polynomials:

<br /> \frac{z^{3} + 1}{z(1 - z^{2})} = \frac{z^{3} + 1}{-z^{3} + z} = \frac{z^{3} - z + z + 1}{-z^{3} + z} = -1 - \frac{z + 1}{z^{3} - z}<br />

Next, the denominator factors into linear factors:
<br /> z^{3} - z = z (z^{2} - 1) = z (z - 1) (z + 1)<br />
so, you have:
<br /> \frac{z + 1}{z^{3} - z} = \frac{z + 1}{z ( z - 1) (z + 1)} = \frac{A}{z} + \frac{B}{z - 1} + \frac{C}{z + 1}<br />
You find A, B and C.

 

[bryansteele]

(z³+1)/z(1-z²)

Do you mean:

(z³+1)/(z(1-z²))

or

((z³+1)/z)(1-z²)

Hey, I made a quick edit in the problem. I meant the quantity squared. Sorry for the confusion.

 

[Dickfore]

Again, your numerator and denominator have the same power. Do the long division:

<br /> (z^3 + 1) \colon (z^3 - 2 z^2 + z) = ?<br />

first. What is the quotient and what is the remainder?

 

[bryansteele]

It comes out to ((z+1)(z²-z+1))/(z(z-1)²)

 

[Dickfore]

no, this is not long division.

 

[bryansteele]

Right. It is 1 + (2z²-z+1)/(z³-2z²+z)

 

[Dickfore]

How?

<br /> 1 \cdot (z^{3} - 2 z^{2} + z) + (2 z^{2} - z) = z^{3} - 2 z^{2} + z + 2 z^{2} - z = z^{3} \neq z^{3} + 1<br />

 

[bryansteele]

right again. I've edited my last reply. I forgot the +1 in the numerator.

 

[Dickfore]

Ok, now look at the normal fraction (power of numerator is lower than power of denominator). The partional fraction decomposition is:

<br /> \frac{2 z^{2} - z + 1}{z ( z - 1)^{2}} = \frac{A}{z} + \frac{B}{z - 1} + \frac{C}{(z - 1)^{2}}<br />

If you multiply by z (z - 1)^{2}, and compare coefficients in front of like powers of z, you will get 3 equations for A, B and C. Solve them and you are done.

 

[bryansteele]

Great, appreciate the patience. Imagine, I've made it to a graduate-level complex analysis class without ever doing partial fractions...

Loads of help, thanks again.

 

[Dickfore]

Cool, now take a look at this:

http://www.wolframalpha.com/input/?i=Apart[(z^3+1)/(z+(z+-+1)^2)]