3
$\begingroup$

I am self-studying complex analysis using Dennis Zill's, A First Course in Complex Analysis. I'm a bit stuck on one of the exercises in Chapter 2.4, which asks us to find the image of the region shown in the figure below under the principal square root function $w=z^\frac{1}{2}$.

enter image description here

My naive attempt, based on the Documentation for ComplexRegionPlot is:

region[z_] := 0 >= Re[z] >= 4 - (Im[z])^2/16 && Im[z] >= 0;
f[z_] := z^(1/2);
{ComplexRegionPlot[region[z], {z, 20}, PlotLabel -> z],
 ComplexRegionPlot[region[InverseFunction[f][z]], {z, 20}, 
  PlotLabel -> f[z]]}

Which gives me:

enter image description here

My confusion is, why am I getting two triangular regions for the image? Shouldn't the image be limited to the single triangular region in the first quadrant?

EDIT: Upon further reflection, it appears that my function f[z_] := z^(1/2) is not the principal square root function. If I instead define a new function,

g[z_] := Sqrt[Abs[z]] E^(I Arg[z]/2)

then do

{ComplexRegionPlot[region[z], {z, 20}, PlotLabel -> z],
 ComplexRegionPlot[region[InverseFunction[g][z]], {z, 20}, 
  PlotLabel -> g[z]]}

then I get the desired result:

enter image description here

Improve this question
$\endgroup$
2
  • $\begingroup$ Perhaps it is because my function f[z_] := z^(1/2) is not the principal square root function but instead is giving both square roots. I will edit my original post accordingly. $\endgroup$
    – bmclaurin
    Commented Mar 14, 2022 at 23:57
  • $\begingroup$ The image (Image) of this this domain form can also be studied with geogabra: have a look there too. geogebra.org/m/gt6YKZnh $\endgroup$
    – janhardo
    Commented Mar 22, 2022 at 10:08

2 Answers 2

Reset to default
4
$\begingroup$

…it appears that my function f[z_] := z^(1/2) is not the principal square root function.

No, actually it's the InverseFunction that's not calculating the inverse function in a mathematically rigorous manner. As mentioned in Possible Issues section of document of InverseFunction:

Equations $f^{(-1)} (f (x))=x$ and $f\left(f^{(-1)}(y)\right)=y$ may not hold for arbitrary $x$ and $y$.

If you use Solve to calculate the inverse, a warning pops up (at least since v12.3.1):

Solve[f[z] == w, z]

Solve::nongen: Solutions may not be valid for all values of parameters.

To calculate the inverse function rigorously, let's add a constraint to Solve:

fz = z /. First@Solve[{f[z] == w, Im[z] > 0}, z]
(* ConditionalExpression[w^2, Re[w] > 0 && Im[w] > 0] *)

ComplexRegionPlot[region[fz], {w, 20}, PlotLabel -> f[z], PlotPoints -> 50]

enter image description here

Improve this answer
$\endgroup$
4
  • $\begingroup$ Very nicely done. I think it's noteworthy that while Solve complains inV13, in V12 it does not. $\endgroup$
    – bmf
    Commented Mar 23, 2022 at 5:41
  • $\begingroup$ @bmf Oh, didn't notice it's a new feature. The warning is already in v12.3.1. $\endgroup$
    – xzczd ♦
    Commented Mar 23, 2022 at 5:57
  • $\begingroup$ thanks for letting me know about this :) $\endgroup$
    – bmf
    Commented Mar 23, 2022 at 5:58
  • $\begingroup$ @xzczd thank you very much for this and apologies for taking so long to get back to it. this will definitely help me going forward. $\endgroup$
    – bmclaurin
    Commented Apr 11, 2022 at 15:49
0
$\begingroup$

Using package :

Needs["Graphics`ComplexMap`"]

can be used Which replacement commands to find for CartesianMap and PolarMap?

Improve this answer
$\endgroup$
2
  • 3
    $\begingroup$ How can this package be used to solve OP's problem? Can you elaborate a bit? $\endgroup$
    – xzczd ♦
    Commented Mar 23, 2022 at 4:32
  • $\begingroup$ I read here for linear mapping : The transformation w = f(z) = z^(1/2) usually maps vertical and horizontal lines onto portions of hyperbolas. I have a few plot examples of linear mappping here that use the package Needs["GraphicsComplexMap`"] How to solve this issue takes some time.... $\endgroup$
    – janhardo
    Commented Mar 23, 2022 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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