Evaluating Residue at pole with different forms

Question

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.

Answer 1

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]))