Riemann surface for cubic root

I'm trying to generate Riemann surfaces of higher order. I read How to visualize Riemann surfaces?, and the subsequent documents. I'm able to reproduce the surface using polar representation, but I'm having a difficult time generating them using rectangular coordinates, for roots greater then 2. So far what I have is:

p1 =
Plot3D[{Im[(x + I*y)^(1/2)], -Im[(x + I*y)^(1/2)]}, {x, -2, 2}, {y, -2, 2},
PlotPoints -> {40, 120}, Mesh -> 25, BoxRatios -> {2, 2, 2},
ColorFunction -> Function[{x, y, z}, Hue[z]]]

p2 =
ParametricPlot3D[
Evaluate[({r Cos[#1], 0, (r)^(1/3) Sin[#1/3]} &) /@ {0, π,2 π, 3 π}],
{r, 0, 2},
PlotStyle -> {{Black, AbsoluteThickness[5]}}]

Show[p1, p2]


Rectangular coordinates and branches in black

p3 =
ParametricPlot3D[{r Cos[φ], r Sin[φ], r^(1/2) Sin[φ/2]}, {r, 0, 1}, {φ, 0, 4 π},
ColorFunction ->
Function[{x, y, z, r, φ, θ}, {Specularity[#], Glow[#]}& @ Hue[Rescale[φ, {0, 1}]]],
PlotPoints -> {20, 60}, Mesh -> 25]

Show[p3, p2]


Polar coordinates with branches in black

The square roots are both identical. If I go on the cubic roots, the polar representation is the same with a change from 2 to 3, but I don't know how to modify the rectangular code to generate the same surface. So for example:

p4 =
ParametricPlot3D[{r Cos[φ], r Sin[φ], (r)^(1/3)Sin[φ/3]}, {r, 0, 1}, {φ, 0, 6 π},
ColorFunction ->
Function[{x, y, z, r, φ, θ}, {Specularity[#], Glow[#]}& @ Hue[Rescale[φ, {0, 1}]]],
PlotPoints -> {40, 120},
Mesh -> 25]

p5 =
ParametricPlot3D[
Evaluate[({r Cos[#1], 0, (r)^(1/3) Sin[#1/3]} &) /@ {0, π,2 π, 3 π, 4π, 5π}],
{r, 0, 2},
PlotStyle -> {{Black, AbsoluteThickness[5]}}]

Show[p4, p5]


Cubic root in polar with branches in black

I have no idea how to get the same cubic root surface using polar coordinates.

I assume when you wrote "I have no idea how to get the same cubic root surface using polar coordinates", you meant rectangular coordinates because you already have the surface in polar coordinates.

In the first plot, p1, you're plotting Im[(x + I y)^(1/2)] and Im[-(x + I y)^(1/2)] because $1$ and $-1$ are the two square roots of unity. To plot cube roots, you just have to use the cube roots of unity, namely $1$, $e^{2\pi i/3}$, and $e^{4\pi i/3}$:

Plot3D[
{Im[(x + I y)^(1/3)],
Im[E^(2 I Pi/3) (x + I y)^(1/3)],
Im[E^(4 I Pi/3) (x + I y)^(1/3)]},
{x, -2, 2}, {y, -2, 2},
PlotPoints -> {40, 120}, Mesh -> 25, BoxRatios -> {2, 2, 2},
ColorFunction -> Function[{x, y, z}, Hue[z]]
]


(I didn't change any of the other options in your plot.)