I am very new to Mathematica and overall not very good with programming. I want to plot the region $|(\frac{z}{f(z)}f'(z))^{1+q}-1|<1$ where $z$ is taken from the unit disk and $0<q<1$. In this example, I took the function $f$ to be $f(z)=z+\frac{1}{2}z^2$ and this is the code I used to draw the region (with the help of ChatGPT admittedly).

  (* Define the function f[z] *)
f[z_] := z+(1/2)z^2

(* Compute the derivative of f[z] with respect to z *)
fDerivative[z_] := D[f[z], z]

(* Define a range of q values from 0 to 1 *)
qValues = Range[0, 1, 0.1];

(* Create an animation of plots for different q values *)
animation = Animate[
    Abs[((z/f[z])^(1 + q) fDerivative[z]) - 1] < 1,
    {z, 3},
    PlotLabel -> Row[{"q = ", q}],
    PlotRange -> All],
  {q, qValues},
  AnimationRunning -> False

I have a feeling that this code is not correct. I want the values of z to be taken from the unit disk $|z|<1$.

Is it as simple as changing the bit of code in ComplexRegionPlot to be like this?

  Abs[((z/f[z])^(1 + q) fDerivative[z]) - 1] < 1 && Abs[z]<1,
    {z, 3}]

Would that work in fixing my problem or do I need to add or change some other parameter to get the results I wanted? Or is it already correct and I am just confused over nothing.

  • 1
    $\begingroup$ Note that your formula is: ((z/f[z] fDerivative[z] )^(1 + q) ) , but your code is: ((z/f[z])^(1 + q) fDerivative[z]) $\endgroup$ Aug 2, 2023 at 8:01
  • $\begingroup$ It is not necessary to define a separate function for the derivative. Just use f'[z]. They are identical (SameQ), i.e., f'[z] === D[f[z], z] is True $\endgroup$
    – Bob Hanlon
    Aug 2, 2023 at 15:10

1 Answer 1


Applying @DanielHuber's useful comment we get

    Abs[ (z/f[z]  fDerivative[z])^(1 + q) - 1] < 1 && Abs[z] < 1, {q, 
     0, 1, .25}], {z, 3}] // Quiet

enter image description here


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