2
$\begingroup$

I'm interested in investigating the following function f[w].

g[z_, b_, r_, a_] := (z - b) / (z - r * I^a) h[z_,b_,r_,a_] := g[z,b,r,a] + g[Conjugate[z],b,r,a] f[w_, b_, r_, a_] := h[I^w,b,r,a]

Function h for given parameters

It's a complex function (h[z]), but we constrain z to the unit circle. I would like to solve a system of equations to find a and b for a given r. I believe this would be expressed as such:

Solve[{Im[f[w,b,r,a]]==0, Re[D[f[w,b,r,a],w]]==0} /. {w->0.2, r->0.97}, {a,b}]

That does not work, but I'm not sure what can be done about it.

I have been able to create real-value functions related to this problem and then solve for {x,b} given r and a.

In[71]:= Solve[{redf[x,b,r,a]==0,0.01<x<0.5,imf[x,b,r,a]==0,0.01<x<0.5}/.{r->0.97,a->0.25},{x,b}]

{"Solve", "ratnz", ""Solve was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.""}

Out[71]= {{x->0.249226,b->-0.247524}}

That's close to what I'm looking for. I'm looking for advice working with this problem:

  1. Is it a good idea to transform all equations to have all variables as real? What's the best way to work with complex equations, and impose constraints such as Abs[x]==0?
  2. How can I figure out if a closed formula for {a,b} given {w,r} might be attainable?
  3. It seems possible to at least solve it numerically, what am I missing?
  4. Is there a better way to deal with the derivative, considering it's a complex function? (More of a theoretical question)
Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Well...there is the Wolfram Function Repository function TrigNSolve intended for such problems. Unfortunately the author kind of glitched and it does not work on the relevant system. I'll show how to do so in an actual response once the needed fixes get republished (maybe tomorrow, I hope). $\endgroup$ Commented Apr 24 at 17:23
  • 2
    $\begingroup$ Also, did your parents name you "dividebyzero"? Didn't anyone teach them not to do that? $\endgroup$ Commented Apr 24 at 17:24

2 Answers 2

Reset to default
2
$\begingroup$

Finally the WFR function TrigNSolve is ready for this system.

I start with the given system.

g[z_, b_, r_, a_] := (z - b)/(z - r*I^a) h[z_, b_, r_, a_] := g[z, b, r, a] + g[Conjugate[z], b, r, a] f[w_, b_, r_, a_] := h[I^w, b, r, a]

Extract real and imaginary parts. Then use Together and Numerator to create a "nice" system. Note that solutions that make the denominator vanish might be invalid; this is something for which I did not check.

{re, im} = ComplexExpand[ReIm@f[w, b, r, a]]; im1 = Numerator[Together[im]]; rederiv1 = Numerator[Together[D[re, w]]]; tpolys = {im1, rederiv1} /. {w -> 1/5, r -> 97/100};

Now solve this.

ResourceFunction["TrigNSolve"][N[tpolys] == 0, {a, b}] (* Out[32]= {{a -> 1.84318, b -> 1.05314}, {a -> -1.84318, b -> 1.05314}, {a -> -0.200965, b -> -0.445669}, {a -> 0.200965, b -> -0.445669}, {a -> 0.2 + 0.0193909 I, b -> 0.951057 - 0.309017 I}, {a -> 0.2 - 0.0193909 I, b -> 0.951057 + 0.309017 I}, {a -> -0.2 - 0.0193909 I,b -> 0.951057 - 0.309017 I}, {a -> -0.2 + 0.0193909 I, b -> 0.951057 + 0.309017 I}, {a -> 0.0156461, b -> 0.988719}, {a -> 0., b -> 0.97}, {a -> 2., b -> -0.97}, {a -> -0.0156461, b -> 0.988719}} *)

I took the liberty of adding this example and a link to the MSE post to the TrigNSolve entry. Also to the Possible Issues section (those who might be curious can check there for the reason).

Improve this answer
$\endgroup$
1
$\begingroup$

To get a solution for a and b, provided w->0.2, r->0.97 we can define a new function:

ff[b_, a_] := f[0.2, b, 0.97, a]

and search for roots of ff:

Reduce[ff[b, a] == 0, {a, b}]

enter image description here

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.