MHB Partial Fraction Decomposition of Complex Fractions

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The correct partial fraction decomposition of the complex fraction \( \frac{1}{z^2(1-z)} \) should include three terms: \( \frac{A}{z} + \frac{B}{z^2} + \frac{C}{1-z} \). The original poster initially omitted the \( \frac{A}{z} \) term, which is necessary for proper decomposition. It is also clarified that the coefficients \( A, B, C \) are assumed to be complex numbers. This discussion emphasizes the importance of including all necessary terms in partial fraction decomposition. Understanding this concept is crucial for accurately solving complex fraction problems.
Dustinsfl
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$$
\frac{1}{z^2(1-z)} = \frac{A}{z^2}+\frac{B}{1-z}
$$I can't figure out how to decompose this fraction.
 
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dwsmith said:
$$
\frac{1}{z^2(1-z)} = \frac{A}{z^2}+\frac{B}{1-z}
$$I can't figure out how to decompose this fraction.

You're missing a term. The decomposition should be of the form

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

Is it also to be assumed that $A,B,C\in\mathbb{C}$?
 
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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