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

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To expand the function f(z)=1/(1+z^3)^2 using partial fractions, it is essential to factor the denominator as (1+z)^2(z^2-z+1)^2. The partial fraction decomposition can then be expressed as A/(z+1) + B/(z+1)^2 + (Cx+D)/(z^2-z+1) + (Ex+F)/(z^2-z+1)^2. The discussion also references a previous thread about using complex coefficients for partial fractions, suggesting a connection to similar problems. Understanding the factorization and applying the correct partial fraction method is crucial for solving this type of problem effectively.
jameson2
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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.
 
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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?
 
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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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