# Plotting surface [closed]

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:

• Can you give more details? Jun 9, 2022 at 10:10
• Please post the details about the original complex equation instead of only the real equation. Jun 9, 2022 at 12:24
• it's posted now. Jun 9, 2022 at 12:41
• @user981656: Do you really mean an indefinite integral? Jun 9, 2022 at 12:45
• I meant in particular the code for the formula for $X(z)$, which you assume is correct. You are more likely to get more people to try to plot it for you if they can copy-paste instead of transcribing from browser to Mma. That said, incorrect code often reveals a misconception that you might like to have cleared up, which is a separate reason for posting code. (The first comment was meant to be generally helpful. You aren't required to post code, but some people will just skip a problem with no code and go on to something easier to deal with. Or go back to work. :) Jun 9, 2022 at 15:39

(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)}


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}]