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Consider the following algebraic expression

-(1/((rm - rp)^2 (-1 + z) z (3 + \[Nu]^2)^2))l^2 (4 k^2 (-1 + z)^2 -4 k (-1 + z) (2 rp \[Nu] - 2 rm z \[Nu] + (-1 + z) Sqrt[rm rp (3 + \[Nu]^2)]) \[Omega] + (-rp + rm z) (3 rp z +rp (-4 + z) \[Nu]^2 -4 (-1 + z) \[Nu] Sqrt[rm rp (3 + \[Nu]^2)] + rm (-3 + (-1 + 4z)\[Nu]^2)) \[Omega]^2)

I am trying to extract the coefficients $A$, $B$, $C$ and $D$ of the above expression, which can be rewritten in the form (through a calculation I did by hand) $A/[z(1-z)]+B/(1-z)+C/z+D$. I tried with the Collect command and different variations of it, but I am only being able to extract $C$.

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  • $\begingroup$ There is an ambiguity in your problem: if $(A,B,C,D)$ satisfies the condition, so does $(A + G, B - G, C - G, D)$ for arbitrary $G$, because $G/((1 - z) z) -G/(1 - z) - G/z = 0$. $\endgroup$
    – A. Kato
    Commented Aug 20 at 6:21

2 Answers 2

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All rationals have internal representation as products with negative powers. Replace

f=Collect[ expr/.{(-1+z)^(-1):> u, z^(-1):> v}, {u,v]] 

Replace complex subexpressions first, order with not independent factors matters.

With

Select[Flatten@Coefficientlist[f,{u,v}],(#=!=0&)]

it is possible to separate all coefficients.

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To make things more clear, remove the overall factor in our expression:

ex = -(1/((rm - rp)^2 (-1 + z) z (3 + \[Nu]^2)^2)) l^2 (4 k^2 (-1 + 
         z)^2 - 4 k (-1 + z) (2 rp \[Nu] - 
        2 rm z \[Nu] + (-1 + z) Sqrt[
          rm rp (3 + \[Nu]^2)]) \[Omega] + (-rp + rm z) (3 rp z + 
        rp (-4 + z) \[Nu]^2 - 
        4 (-1 + z) \[Nu] Sqrt[rm rp (3 + \[Nu]^2)] + 
        rm (-3 + (-1 + 4 z) \[Nu]^2)) \[Omega]^2);

by multiplying:

 ex ((rm - rp)^2 (-1 + z) z (3 + \[Nu]^2)^2)

This gives:

enter image description here

As you can see, there are terms with (-1+z), (-1+x)^2, (-1+z)z, (-4+z)z, but no inverse terms.

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