Plotting surface [closed]

Question

Does anyone know how to plot this image? I am writing an example for Kenmotsu's representation theorem , for $\varphi :% %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion -\{0\}\rightarrow %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion $, $\varphi (z)=\frac{1}{\bar{z}^{2}}.$ And this image corresponds to the surface with $\varphi (z)=-\frac{1}{\bar{z}^{2}}$ and $% H=1$. I am using Mathematica for only a couple days and I don't know what I'm supposed to write to generate it. I appreciate any advice and help! Thank you!

after a fairly long calculation I get to:

and I assume that using this equation plots the image but I don't realize how I need to write this in Mathematica.

complex equation:

Answer 1

(This is too long for a comment.)

If you look at the result of

(Re[f[u + I v]] + I Im[f[u + I v]]) (\[DifferentialD] u + I \[DifferentialD] v) // Expand
   I \[DifferentialD]u Im[f[u + I v]] - \[DifferentialD]v Im[f[u + I v]] +
   \[DifferentialD]u Re[f[u + I v]] + I \[DifferentialD]v Re[f[u + I v]]

this gives a hint on how to compute $\Re\int f(z,\bar{z})\,\mathrm dz$:

parts = Simplify[ComplexExpand[ReIm[1/z^3 {(1 - Conjugate[z]^-4)/(1 + (z Conjugate[z])^-2)^2,
                                           I (1 + Conjugate[z]^-4)/(1 + (z Conjugate[z])^-2)^2,
                                           (2 Conjugate[z]^-2)/(1 + (z Conjugate[z])^-2)^2} /.
                                    z -> u + I v], TargetFunctions -> {Re, Im}]]
   {{(u (-1 + u^4 - 2 u^2 v^2 - 3 v^4))/(1 + u^4 + 2 u^2 v^2 + v^4)^2,
     (v (-1 - 3 u^4 - 2 u^2 v^2 + v^4))/(1 + u^4 + 2 u^2 v^2 + v^4)^2},
    {-((v (1 - 3 u^4 - 2 u^2 v^2 + v^4))/(1 + u^4 + 2 u^2 v^2 + v^4)^2),
     (u + u^5 - 2 u^3 v^2 - 3 u v^4)/(1 + u^4 + 2 u^2 v^2 + v^4)^2},
    {(2 u (u^2 + v^2))/(1 + u^4 + 2 u^2 v^2 + v^4)^2,
     -((2 v (u^2 + v^2))/(1 + u^4 + 2 u^2 v^2 + v^4)^2)}}

FullSimplify[(MapThread[Integrate, {#, {u, v}}] & /@ parts) . {1, -1}]
   {(-u^2 + v^2)/(1 + (u^2 + v^2)^2), -((2 u v)/(1 + (u^2 + v^2)^2)), -(1/(1 + (u^2 + v^2)^2))}

which you should then multiply by a factor of $\frac2{H}$.

Verify the constant curvature property:

Simplify[ResourceFunction["MeanCurvature"][2 {(-u^2 + v^2)/(1 + (u^2 + v^2)^2),
                                              -((2 u v)/(1 + (u^2 + v^2)^2)),
                                              -(1/(1 + (u^2 + v^2)^2))}, {u, v}],
         u ∈ Reals && v ∈ Reals]
   1

As an exercise, why is the following result expected?

GroebnerBasis[Thread[{x, y, z} == 2 {(-u^2 + v^2)/(1 + (u^2 + v^2)^2),
                                     -((2 u v)/(1 + (u^2 + v^2)^2)),
                                     -(1/(1 + (u^2 + v^2)^2))}],
              {x, y, z}, {u, v}] // Simplify
   {x^2 + y^2 + z (2 + z)}

Answer 2

This (?minimal) surface in three dimensions can be plotted in such ways:

ParametricPlot3D[{-(x^2 - y^2)*(1 - (x^2 + y^2)^2)/(1 + (x^2 + y^2)^2), 
-2*x*y*(1 - (x^2 + y^2)^2)/(1 + (x^2 + y^2)^2), 
(x^2 +  y^2)^2/(1 + (x^2 + y^2)^2)}, {x, -1, 1}, {y, -1, 1}]

or switching to the polar coordinates

ParametricPlot3D[{-(x^2 - y^2)*(1 - (x^2 + y^2)^2)/(1 + (x^2 + y^2)^2), -2*x*
y*(1 - (x^2 + y^2)^2)/(1 + (x^2 + y^2)^2), (x^2 +  y^2)^2/(1 + (x^2 + y^2)^2)} 
/. {x -> r*Cos[t], y -> r*Sin[t]}, {r, 0, 3/2}, {t, 0, 2*Pi}]

See the documentation for more info.