Partial fraction decomposition

  • #1
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!
 
Last edited:

Answers and Replies

  • #2
353
1
(z3+1)/z(1-z2)

Do you mean:

(z3+1)/(z(1-z2))

or

((z3+1)/z)(1-z2)
 
  • #3
2,967
5
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:

[tex]
\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}
[/tex]

Next, the denominator factors into linear factors:
[tex]
z^{3} - z = z (z^{2} - 1) = z (z - 1) (z + 1)
[/tex]
so, you have:
[tex]
\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}
[/tex]
You find A, B and C.
 
  • #4
(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.
 
  • #5
2,967
5
Again, your numerator and denominator have the same power. Do the long division:

[tex]
(z^3 + 1) \colon (z^3 - 2 z^2 + z) = ?
[/tex]

first. What is the quotient and what is the remainder?
 
  • #6
It comes out to ((z+1)(z2-z+1))/(z(z-1)2)
 
  • #7
2,967
5
no, this is not long division.
 
  • #8
Right. It is 1 + (2z2-z+1)/(z3-2z2+z)
 
Last edited:
  • #9
2,967
5
How?

[tex]
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
[/tex]
 
  • #10
right again. I've edited my last reply. I forgot the +1 in the numerator.
 
  • #11
2,967
5
Ok, now look at the normal fraction (power of numerator is lower than power of denominator). The partional fraction decomposition is:

[tex]
\frac{2 z^{2} - z + 1}{z ( z - 1)^{2}} = \frac{A}{z} + \frac{B}{z - 1} + \frac{C}{(z - 1)^{2}}
[/tex]

If you multiply by [itex]z (z - 1)^{2}[/itex], 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.
 
  • #12
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.
 

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