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Find the real and imaginary roots for a complex polynomial?

Posted 4 years ago

Sir, I've a complex polynomial of 7th order of Omega. I want to substitute Omega=OmegaR + i Omega i and evaluate the same for all powers of Omega using binomial theorem. My ultimate goal is to evaluate the real and imaginary roots of the complex polynomial. But I'm getting some errors. Kindly guide me as regarding how should I proceed.

Attached the file for your reference plz.

Sincerely,

Rahul Chakraborty

Attachments:
POSTED BY: Rahul Chakrabory
7 Replies

Yes Sir, I have a complex binomial series, when expanded. Let me try to implement your advice.

It seems the Binomial series that I calculated manually can't get generated by the inbuilt Binomial function. when I calculated manually I substituted

Omega = Omega_R + i Omega_i

I got

Omega^2=Omega_R^2[1+4IOmega_i - 2i Omega_i Omega_R-3Omega_i^2+2Omega_i^2 Omega_R]

I'm unable to generate this series in Mathematica. not sure which function can generate this series.

Sincerely,

POSTED BY: Rahul Chakrabory

The attached notebook is riddled with typographic issues. White space is important. Using spaces where you should not, or failing to use them where you should, will render code unusable. Also notice that the two pattern objects below are different in terms of what they will do.

Subscript[\[CapitalOmega], r] _Integer
Subscript[\[CapitalOmega], i_] Integer
POSTED BY: Daniel Lichtblau
Anonymous User
Anonymous User
Posted 4 years ago

One Eulerian identity of many is that if x is any real number (and under usual other restrictions on the remaining)

( Cos[n x] + I Sin[n x] )^n == ( Cos[n x] + I Sin[n x]) == E^(I n x)

Using the above formula (with crafty substitutions) could lead to brief solution in E^(i n x), compared to the binomial theorem.

Offering spurious theory. In general if a complex function f[z] is defined in Omega it is analytic (holomorphic) in Omega. All real polynomial are analytic. Such an f[z] has a cauchy formula in Omega; roughly equivalent to the taylor polynomial.

POSTED BY: Anonymous User
Posted 4 years ago

I will try to take this a little further.

This gets a clean answer for c==7, z==7.

Solve[(x + y)^z == 
x^c + Sum[Binomial[c, n] x^(c - n) y^(n), {n, 1, c}], z, Reals]

Solve[(x + 1)^z == Sum[Binomial[c, n] x^(n), {n, 0, c}], z, Reals]

The result of this is more complex, but is found by Mathematica: z is a long equation when c==7, mathematica warns inverse functions may not show all solutions...

Solve[(x + y)^z == 
x^c + Sum[Binomial[c, n] x^(c - n) y^(n), {n, 1, c}], z, Complexes]

Solve[(x + 1)^z == Sum[Binomial[c, n] x^(n), {n, 0, c}], z, Complexes]

I'm sure you may have a complex binomial theorem - and one does not immediately jump to my mind other than DeMoirve's, (r(Cos[th]+I Sin[th])^n == r^n(Cos[n th]+I Sin[n th]). (which worker together with euler imaginary identities as well). This shows if your variable "x" can be expressed in terms f[x]==Cos[x] by parameterization (which seems reasonable), then there are binomial theorems that apply that do not apply to Real numbers.

You are perhaps far ahead of me on study of the Complex Plane so I won't continue. But different theorems may be needed depending on the number of variables, whether the exponent is limited in sign or itself or if the exponent may be real and and itself complex. What the objective (in learning) is makes a difference.

https://mathworld.wolfram.com/BinomialTheorem.html

POSTED BY: Updating Name
Anonymous User
Anonymous User
Posted 4 years ago

Solve[expr,vars] assumes by default that quantities appearing algebraically in inequalities are real, while all other quantities are complex.

Solve[expr&&vars[Element]Reals,vars,Complexes] solves for real values of variables, but function values are allowed to be complex.

Solve[(z^7 - 1) (2 z^7 - 4) == 0, z, Complexes]

As to your Binomial theorem I still don't know which theorem(s) you are trying to utilize, which i had asked you identify. However I feel that there are theorems covering 7th order complex polynomial built-in to Mathematica, which was displayed in various ways.

POSTED BY: Anonymous User

Sir, Thanks a lot, but I'm having some doubts, I can't understand some parts of the code, In the above code of yours,

1) In the second part of the code, I could not understand the logic how {z +}), {z, -2 - 2 I, 2 + 2 I} is written. Sir, kindly elaborate for me plz. 2) I,ve tried to implement the corrections as per your guidance, but still there exists some error. kindly guide me. The code is as attached herewith.

Sincerely.

Attachments:
POSTED BY: Rahul Chakrabory
Anonymous User
Anonymous User
Posted 4 years ago
In[177]:= 
Re[z]^7 (1 + 14 I Im[z] - 63 Im[z]^2 - 7 I Im[z] Re[z] + 
    42 Im[z]^2 Re[z]) /. z -> (1 + I)

Out[177]= -20 + 7 I

ComplexPlot[(Re[
     z]^7 (1 + 14 I Im[z] - 63 Im[z]^2 - 7 I Im[z] Re[z] + 
      42 Im[z]^2 Re[z]) /. z -> {z +}), {z, -2 - 2 I, 2 + 2 I}]

this next snip is from Examples in Help for Binomial

ComplexPlot3D[Binomial[z, 1/5], {z, -2 - 2 I, 2 + 2 I}, 
 PlotLegends -> Automatic]

MathematicalFunctionData["Binomial", "NamedIdentities"] // Length == 18
{"Central binomial coefficient", "Eulerian number", ...}

(Mathematica has 18 Binomial theorems which can be shown, if you would like to see them)

I'm unsure about your objective. I don't doubt you have a modified binomial theorem that will apply for complex numbers - there are many theorems to appeal to. But I'm unsure if you are attempting to apply the theorem to find coefficients of the equations (the coefficients are already provided?). I'm unsure if the 7 equations are meant to be combined to prove a binomial theorem.

The notebook contains many syntax errors. You will need to use Help and follow examples or post the text of the theorem you wish to use.

I provided an example plot from the 7th order equation in the notebook.

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POSTED BY: Anonymous User
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