Evaluating Residue at pole with different forms

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


1

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)))

0

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

• Assuming[r > 0, Solve[Denominator[fun] == 0]] yields both solutions. Apr 4, 2018 at 13:08
• @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. Apr 4, 2018 at 13:29

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]))

• 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? Apr 4, 2018 at 15:06
• @TalArsenyMiller Looks like you might be using M10 or earlier? If you use M11 it should work. Apr 4, 2018 at 15:59
• 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. Apr 4, 2018 at 16:16
• @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. Apr 4, 2018 at 16:40
• 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? Apr 4, 2018 at 19:23