Partial Fractions: Expanding 1/(1+z^3)^2

Click For Summary
SUMMARY

The discussion focuses on expanding the function f(z) = 1/(1+z^3)^2 using partial fractions. The correct approach involves factoring the denominator as (1+z)^2(z^2-z+1)^2 and applying partial fraction decomposition. The resulting expression includes terms A/(z+1), B/(z+1)^2, Cx+D/(z^2-z+1), and Ex+F/(z^2-z+1)^2. The conversation also references a previous discussion on the principal part and Laurent series related to 1/(z^2+1)^2.

PREREQUISITES
  • Understanding of partial fraction decomposition
  • Familiarity with polynomial factorization
  • Knowledge of complex roots and their application
  • Basic concepts of Laurent series
NEXT STEPS
  • Study the method of partial fraction decomposition in detail
  • Learn about polynomial factorization techniques
  • Explore the use of complex roots in algebraic expressions
  • Investigate the properties and applications of Laurent series
USEFUL FOR

Mathematics students, educators, and anyone involved in advanced algebra or complex analysis who seeks to deepen their understanding of partial fractions and polynomial expansions.

jameson2
Messages
42
Reaction score
0
I'm trying to do a question that requires the expansion of the following using partial fractions:
f(z)=\frac{1}{(1+z^3)^2}.
The fact that the bottom is squared is throwing me off for some reason... I've factorized the bottom, but I'm not sure whether I should use the complex roots or not, or even if it's possible without using complex roots.
Any help would be appreciated.
 
Physics news on Phys.org
Hey there,
The trick is to use identity 1+Z^3=(1+z)(1-Z+Z^2) and then just square this expression and do the partial fraction as usual.
Hope this helps:)))
 
\frac{1}{(1+z^3)^2}= \frac{1}{(z+1)^2(z^2-z+1)^2}
and can be written as "partial fractions" as
\frac{A}{z+1}+ \frac{B}{(z+1)^2}+ \frac{Cx+D}{z^2- z+1}+ \frac{Ex+F}{(z^2- z+1)^2}

There was an earlier question about the "principal part" or "Laurent series" for 1/(z^2+1)^2 which I answered by reducing to partial fractions with complex coefficients. Is this related to that thread?
 
Last edited by a moderator:

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K