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I want to evaluate residues at the poles of the function $\frac{1}{z^{3/2}+r^{3/2}}$

fun = 1/ (z^(3/2) + r^(3/2));

where z is the variable, and r is a real and positive parameter.

Analytically, there are 2 poles at $z = e^{\pm 2/3 \pi i} r$.

Side Problem:

When I solve for the roots of the denominator, I only get one of the solutions above:

Solve[Denominator[fun] == 0, z]

{{z -> (-r^(3/2))^(2/3)}}

This can be checked to be indeed the solution above with the plus sign:

(-r^(3/2))^(2/3)/(E^(2/3 π I) r) // Simplify

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Any idea why Solve did not find both solutions? Can I "help" it in some way to find both?

Main Problem:

Evaluating the residue using Residue only accepts the form of the solution given by Solve:

Residue[fun, {z, (-r^(3/2))^(2/3)}]
Residue[fun, {z, E^(2/3 π I) r}]

-((2 (-r^(3/2))^(2/3))/(3 r^(3/2)))

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How do I "convince" Mathematica to accept my form of the pole? Or am I wrong in some way? Thanks.

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  • $\begingroup$ Assuming[r > 0, Solve[Denominator[fun] == 0]] yields both solutions. $\endgroup$ Apr 4, 2018 at 13:08
  • $\begingroup$ @AccidentalFourierTransform , actually, even without the "Assuming", Solve[Denominator[fun] == 0] yields both, but Solve[Denominator[fun] == 0, z] doesnt. This is strange, but I do want to specify which variable I'm solving for. In a different case, it could have solved for r instead. $\endgroup$ Apr 4, 2018 at 13:29

1 Answer 1

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Although not documented, Residue does take the Assumptions option:

Options[Residue]

{Assumptions :> $Assumptions}

If you use Assumptions, then Residue is able to give the desired result:

Residue[fun, {z, E^(2/3 π I) r}, Assumptions->r>0]

-((2 E^((2 I π)/3))/(3 Sqrt[r]))

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  • $\begingroup$ The lines fun = 1/(z^(3/2) + r^(3/2)); Residue[fun, {z, E^(2/3 π I) r}, Assumptions->r>0] give 0 nonetheless for me. Any idea what I did wrong? $\endgroup$ Apr 4, 2018 at 15:06
  • $\begingroup$ @TalArsenyMiller Looks like you might be using M10 or earlier? If you use M11 it should work. $\endgroup$
    – Carl Woll
    Apr 4, 2018 at 15:59
  • $\begingroup$ I have M11. I asked a friend that also has M11 to run that line, also got 0. I also noticed that if I first give a numeric value to r, for example r=1/2, I get a non-zero result. On the other hand r=0.5 gives zero again. $\endgroup$ Apr 4, 2018 at 16:16
  • $\begingroup$ @TalArsenyMiller Which version of M11? It works for 11.1, 11.2, and 11.3 for me (on OSX but that shouldn't make a difference). Also, I'm not surprised that Residue with an inexact number produces 0. $\endgroup$
    – Carl Woll
    Apr 4, 2018 at 16:40
  • $\begingroup$ The problem I had was indeed on 11.0, switched to 11.2 and problem disappeared. Thanks alot. What do you think about my "side problem", and the comment above given by AccidentalFourierTransform? $\endgroup$ Apr 4, 2018 at 19:23

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