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i have the following problem: My function is given by:

$(cz^{-1}+aq^{-j}-bq^{-2j}z)\prod_{m=1}^{Q-1}\frac{qz-z_{m}}{z-z_m}+(-cz^{-1}+dq^{j}+bq^{2j}z)\prod_{m=1}^{Q-1}\frac{q^{-1}z-z_{m}}{z-z_m}=E$, where $\lbrace a,b,c,d\rbrace$ are some real coefficients, $q=\exp(i\pi P/Q)$ is some parameter depending on $P,Q$ and $j$ is some integer or half integer

Thus, there are some poles, at $z=0$, $z=\infty$ and at $z=z_{m}$. That is what i did till now:I defined the functions $a(z)$ and $d(z)$ and the products separated from each other. Then i defined a function $K$ to bring them together and in the end I told Mathematica to calculate one of the residues, i.e. at $z=z_m$.

ClearAll[a, b, c, d, q, Q]
a[z_] = c/z + a*q^(j) - bq^(-2 j)*z;
d[z_] = -c/z + d*q^(j) + bq^(2 j)*z;
pr[Q_] = Product[(qz - Subscript[z, m])/(z - Subscript[z, m]), {m, 1, 
Q - 1}];
pr1[Q_] = 
Product[(q^(-1) z - Subscript[z, m])/(z - Subscript[z, m]), {m, 1, 
Q - 1}];
K = a[z]*pr[Q] + d[z]*pr1[Q];
Residue[K, {z, Subscript[z, Infinity]}];

As a result it is:

Residue[K,{z,infinity}]

Here arrives the problem: Mathematica does not solve it? Does anybody has an idea, how i can improve the code to get solutions?

Thanks a lot

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  • $\begingroup$ It looks like you have a typo in the definition of $z_\infty$ - note the lowercase i in the output $\endgroup$ – Lukas Lang Sep 18 '18 at 11:05
  • $\begingroup$ Thanks for your answer. What exactly do you mean? $\endgroup$ – Herrmann Dirac-Neumann Sep 18 '18 at 11:15
  • $\begingroup$ Sorry, I think I've misread your last line of code (which seems to have a typo, i.e. infinity instead of Infinity). The issue is probably that Mathematica cannot compute the answer from the information given. For example, you do not specify the $z_m$ or their relation to each other. $\endgroup$ – Lukas Lang Sep 18 '18 at 11:29
  • $\begingroup$ Okay, but how could I improve the code? I am very sorry, I almost have no experience with Mathematica... $\endgroup$ – Herrmann Dirac-Neumann Sep 18 '18 at 12:13

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