3
$\begingroup$

Let's say I want to plot a branch of the following function $f(z) = \sqrt{z(z-1)}$. Let $z = r_1 e^{i\theta_1}$ and $z-1 = r_2 e^{i\theta_2}$.

How can I tell Mathematica to plot the branch for which $0 \leq \theta_1 < 2\pi$ and $-\pi \leq \theta_2 < \pi$?

$\endgroup$
  • $\begingroup$ Welcome to Mathematica StackExchange. In order to learn how to use this site take the tour. When copying equations from a notebook to your question one should format using inline code by selecting the code and clicking the {} button above the edit window. It is recommended that you browse the Markdown help $\endgroup$ – Jack LaVigne Nov 22 '17 at 22:30
  • $\begingroup$ What sort of plot do you want? $\endgroup$ – bbgodfrey Nov 23 '17 at 3:15
  • $\begingroup$ I would like a 3D plot :) $\endgroup$ – Dory Nov 23 '17 at 8:02
0
$\begingroup$

In a first step you can eliminate r1 ,r2:

erg = 
  Simplify[
    ExpToTrig[Solve[1 - r2 Exp[I θ2] == r1 Exp[I θ1], Element[{r1, r2}, Reals]]], 
    {Element[{θ1, θ2}, Reals], 0 <= θ1 < 2 Pi, -Pi <= θ2 < Pi}][[1]]
{r1 -> -Csc[θ1 - θ2] Sin[θ2], r2 -> Csc[θ1 - θ2] Sin[θ1]}

With this result f[z] evaluates to

Csc[θ1 - θ2]^2 Sin[θ1] Sin[θ2] (Cos[θ1 + θ2] + I Sin[θ1 + θ2])   

and can be plotted in the parameter space of θ1 and θ2.

0 down vote

If x,y depend on teta1, teta2 you can use the function ParametricPlot3D[] with the two parameters teta1, teta2 as arguments to plot f[x+i y].

Sorry, I(as a new user...) don't know why, but I'm not allowed to comment your notes. Though I have to edit my answer:

You're right with the correction. The two surfaces in your plot probably are the real and imaginary part of f. You've to specify the output in your plot , for example Abs[f]!

$\endgroup$
  • $\begingroup$ How can I plot it in the space $(x,y,w)$ with $w=f(z)=f(x+iy)$? $\endgroup$ – Dory Nov 23 '17 at 19:28
  • $\begingroup$ 1) In your Solve function, I think it should be r2 Exp[I θ2] + 1== r1 Exp[I θ1] 2) I tried to plot the function using ParametricPlot3D but it gives me an multiple surfaces although it should give me only one branch of the function : ParametricPlot3D[{-Csc[θ1 - θ2] Sin[θ2] Cos[θ1], -Csc[θ1 - θ2] Sin[θ2] Sin[θ1], Re[f[-Csc[θ1 - θ2] Sin[θ2], -Csc[θ1 - θ2] Sin[θ1], θ1, θ2]]}, {θ1, 0, 2 Pi}, {θ2, -Pi, Pi}] with f[r1_, r2_, θ1_, θ2_] = Sqrt[r1*r2]*Exp[I *(θ1+θ2)/2] $\endgroup$ – Dory Nov 24 '17 at 10:31
  • $\begingroup$ Note : I used the fact that x = r1 Cos[θ1] = -Csc[θ1 - θ2] Sin[θ2] Cos[θ1] and y = r1 Sin[θ1] = -Csc[θ1 - θ2] Sin[θ2] Sin[θ1] for the inputs of the ParametricPlot3D function. $\endgroup$ – Dory Nov 24 '17 at 10:46
0
$\begingroup$

I'm not quite clear what you want but as a first stage why not just plot the modulus (which will be continuous) and the argument (phase) which will show jumps.

   Plot3D[Evaluate[Abs[Sqrt[z (z - 1)]] /. z -> x + I y], {x, -2, 
      2}, {y, -2, 2}]
    Plot3D[Evaluate[Arg[Sqrt[z (z - 1)]] /. z -> x + I y], {x, -2, 
      2}, {y, -2, 2}]

Mathematica graphics

Mathematica graphics

What exactly do you wish to do? Do you wish to have the plot continued beyond the jump edges?

Hope that helps.

$\endgroup$
0
$\begingroup$

To visualize the Riemann surface of $\sqrt{z(z-1)}$ over the OP's chosen domain, do a preliminary analysis with Reduce[]:

Reduce[-π <= Arg[r E^(I θ) - 1] < π && 0 <= θ < 2 π && r >= 0] // FullSimplify
   (r > 0 && (π < θ < 2 π || 0 < θ < π)) || (r == 1 && θ == 0) ||
   (r > 1 && (0 <= θ < ArcSec[r] || (θ + ArcSec[r] > 2 π && θ < 2 π)))

whose result you can then use to construct the RegionFunction to be used in ParametricPlot3D[]:

ParametricPlot3D[{r Cos[θ], r Sin[θ], Im[Sqrt[r Exp[I θ] (r Exp[I θ] - 1)]]},
                 {r, 0, 4}, {θ, 0, 2 π}, 
                 RegionFunction -> Function[{x, y, z, r, θ},
                                            (r > 0 &&
                                             (0 < θ < π || π < θ < 2 π)) ||
                                            (r > 1 &&
                                             (0 <= θ < ArcSec[r] ||
                                              2 π - ArcSec[r] < θ < 2 π))]]

Riemann surface of \sqrt{z(z-1)}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.