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I try to calculate the following residue

Clear[α];
f[z_]:=π^2 Cot[(π z α)^(1/2)];
g[z_]:=f[z+(-π/2 + 2 k π)^2/(π α)];
res[k_]=Residue[FullSimplify[TrigToExp@g[z],k\[Element]Integers],{z,0}]

but it does not work. Is it possible calculate the residue automatically?

To clarify: if you try to calculate

f[z_]:= π^2 Tan[(π z α)^(1/2)];
Assuming[k\[Element]Integers,Residue[f[z],{z,(π/2+k π)^2/(π α)}]]

you get $0$, but if you calculate

f[z_]:= π^2 Tan[(π z α)^(1/2)];
Assuming[k\[Element]Integers,Residue[f[z],{z,(π/2+1 π)^2/(π α)}]]

you get -((3 π^2)/α)

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  • 1
    $\begingroup$ Once I corrected your syntax by inserting several semicolons to separate lines and ran the code I got $0$... Might this be the correct answer? $\endgroup$ – David G. Stork Dec 7 '17 at 18:55
  • $\begingroup$ No if you subtitute 1 in k no gives answer $\endgroup$ – antonio asis Dec 7 '17 at 18:57
  • $\begingroup$ I get res[1] = 0. Please fix your syntax and re-run your code. And why is your title Residue4? $\endgroup$ – David G. Stork Dec 7 '17 at 18:58
  • $\begingroup$ David Please look the question $\endgroup$ – antonio asis Dec 7 '17 at 19:12
  • $\begingroup$ And what is wrong with those answers? What exactly is the problem you're facing? $\endgroup$ – David G. Stork Dec 7 '17 at 19:31
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One possibility is to avoid dealing with the singularity by considering the inverse:

f[z_] := π^2 Tan[(π z α)^(1/2)];

inv = Simplify[
    Series[1/f[z], {z, (π/2+k π)^2/(π α), 1}],
    Assumptions->k ∈ Integers && k>0
];
inv //TeXForm

$-\frac{\alpha \left(z-\frac{\pi (2 k+1)^2}{4 \alpha }\right)}{\pi ^2 (2 k+1)}+O\left(\left(z-\frac{\pi (2 k+1)^2}{4 \alpha }\right)^2\right)$

Now we can find the residue of the inverse of the inverse:

res[k_] = Residue[1/inv, {z, (π/2+k π)^2/(π α)}];
res[k]

-(((1 + 2 k) π^2)/α)

Let's compare:

res /@ Range[5]
Residue[f[z], {z, (π/2+# π)^2/(π α)}]& /@ Range[5]

{-((3 π^2)/α), -((5 π^2)/α), -(( 7 π^2)/α), -((9 π^2)/α), -((11 π^2)/α)}

{-((3 π^2)/α), -((5 π^2)/α), -(( 7 π^2)/α), -((9 π^2)/α), -((11 π^2)/α)}

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  • $\begingroup$ Thanks Carl Woll again $\endgroup$ – antonio asis Dec 7 '17 at 21:31
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The same solution using Tableand FindSequenceFunction:

f[z_] := π^2*Tan[(π z α)^(1/2)]
sol = Table[Residue[f[z], {z, ((π/2) + k π)^2/(π α)}], {k, 1, 5}]
Res = FindSequenceFunction[sol, k] // Simplify

$\left\{-\frac{3 \pi ^2}{\alpha },-\frac{5 \pi ^2}{\alpha },-\frac{7 \pi ^2}{\alpha },-\frac{9 \pi ^2}{\alpha },-\frac{11 \pi ^2}{\alpha }\right\}$

$-\frac{\pi ^2 (2 k+1)}{\alpha }$

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