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I try to calculate f[z_] := Coth[z]/( E^z - 1)^1 and calculate

Reduce[(f[z])^-1 == 0, z, GeneratedParameters -> (k &)] // FullSimplify

and gives 

k \[Element] Integers && (z == 2 I k \[Pi] || z == I k \[Pi])

now i try to pass the poles to the residue to calculate

Table[Residue[f[z],{z,k1}],{k1,{2I \[Pi] k,-I \[Pi] k,-2I \[Pi] k,I \[Pi] k}}]

it is possible to backsustitution in the table above automatically thanks anyway

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like this?

sol = FullSimplify@Assuming[C[1] \[Element] Integers, 
Simplify@Solve[Simplify@Reduce[ComplexExpand@(f[z]^-1) == 0, z], z]]

Assuming[ C[1] \[Element] Integers,FullSimplify@Residue[f[z], Evaluate@{z, z /. sol[[1]]}]]

(* -1/2 *)

I think that here it is assumed it is known C[1] is an integer. So, I suggest to look at what Reduce returns to be sure.

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  • $\begingroup$ Jose Antonio how to calculate the result when k is negative or positive apart $\endgroup$
    – antonio asis
    Commented Nov 18, 2017 at 18:12
  • $\begingroup$ But the solution includes all the integers isn't it? So the result is valid for negative as well as positive integers. I think you could use C[1] \[Element] Integers && C[1] > 0 or C[1] \[Element] Integers && C[1] < 0 in Assuming $\endgroup$
    – José Antonio Díaz Navas
    Commented Nov 18, 2017 at 18:16
  • $\begingroup$ Jose alreadyy test your solution but the Residue at poles when k is positive is not the same at k negative reduce not discriminate the solution when k is posivite the pole the residue y bot the same in z=2k i that in z=.-2 k i it is posible to get thos to solution apart above but i need that reduce separate C[1] in interger + and - thanks anyway $\endgroup$
    – antonio asis
    Commented Nov 18, 2017 at 18:29
  • $\begingroup$ $$\frac{\csc (x)}{e^{\pi x}-1}=-\sum _{k=1}^{\infty } \frac{(-1)^k x^2}{\pi \left(e^{\pi ^2 k}-1\right) k (\pi k-x) (\pi k+x)}-\frac{(-1)^k x^3}{\pi ^2 \left(e^{\pi ^2 k}-1\right) k^2 (\pi k-x) (\pi k+x)}+\frac{(-1)^k e^{\pi ^2 k} x^3}{\pi ^2 \left(e^{\pi ^2 k}-1\right) k^2 (\pi k-x) (\pi k+x)}-\frac{x^2 \text{csch}(k) \text{sech}(k)}{2 \pi k \left(4 k^2+x^2\right)}-\frac{(-1)^k e^{\pi ^2 k} x^2}{\pi \left(e^{\pi ^2 k}-1\right) k (\pi k-x) (\pi k+x)}+\frac{1}{\pi x^2}-\frac{x}{12}-\frac{1}{2 x}+\frac{2+\pi ^2}{12 \pi }$$ reduce not separate C[1] + or - i need to series as above $\endgroup$
    – antonio asis
    Commented Nov 18, 2017 at 18:32
  • $\begingroup$ I think the result is correct as the MacLaurin Series gives the value $-1/2$ for the coefficient of $1/(z-z_0)$ independently of the sign of the integer C[1] $\endgroup$
    – José Antonio Díaz Navas
    Commented Nov 18, 2017 at 18:41

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