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Hi can you suggest me a way to Reduce/solve the following equation(analytic/numeric):

eqn= {-a1 + I (\[Pi] (((1/2 - I) + \[Pi]/3) Conjugate[
       r1] + ((1/2 + I) + \[Pi]/3) Conjugate[r2]) + 
  3 \[Pi] (((1 - 2 I) + (2 \[Pi])/3) Conjugate[
       r1] + ((1 + 2 I) + (2 \[Pi])/3) Conjugate[r2])) == 0, -a2 + I (\[Pi] (Conjugate[
       r1] ((-I + Sqrt[3]/2)^2 + Cos[(2 \[Pi])/9]) + 
     Conjugate[r2] ((I + Sqrt[3]/2)^2 + Cos[(2 \[Pi])/9])) + 
  3 \[Pi] (Conjugate[
       r1] ((-2 I + Sqrt[3])^2 + 2 Cos[(2 \[Pi])/9]) + 
     Conjugate[r2] ((2 I + Sqrt[3])^2 + 2 Cos[(2 \[Pi])/9]))) == 0, C2 (1 - n1) - C0 (1 + n1) - 4 Im[(A a1 + A1 a2) r1] == 0, -C0 (1 + n2) - 4 Im[(a2 B + a1 B1) r2] == 0, -r1 + I n1 Conjugate[A a1 + a2 B1] == 0, -r2 + I n2 Conjugate[a1 A1 + a2 B] == 0};

var={a1, a2, r1, r2, n1, n2};
Simplify[Reduce[eqn, var]];

It's running for a long time. Thanks.

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  • $\begingroup$ Break it down. You can solve for Conjugate[r1] and Conjugate[r2] from your first two equations. To go further you will need assumptions for your known variables a1,a2,A1,A1 etc. or input the values if you have them. I cannot tell if they are real or complex and MMA can't either. $\endgroup$
    – Bill Watts
    Commented May 3, 2019 at 22:17
  • $\begingroup$ @BillWatts, Thanks bill. All variables are unknown and complexes except for n1 and n2. Though, I'll try. $\endgroup$
    – Rupesh
    Commented May 4, 2019 at 3:02

1 Answer 1

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[Not a full answer but might all the same be of use.]

An analytic solution might be out of reach. I would suggest as a start the following.

(1) Recast in terms of explicit real variables using ComplexExpand. (2) Convert trig constants to radicals to make algebraic dependencies more obvious. (3) Clear denominators to obtain explicit polynomials.

These can be done as below.

exprs = Numerator[
  Together[ToRadicals[
    Flatten[ComplexExpand[
       ReIm[{-a1 + 
           I (\[Pi] (((1/2 - I) + \[Pi]/3) Conjugate[
                   r1] + ((1/2 + I) + \[Pi]/3) Conjugate[r2]) + 
              3 \[Pi] (((1 - 2 I) + (2 \[Pi])/3) Conjugate[
                   r1] + ((1 + 2 I) + (2 \[Pi])/3) Conjugate[
                   r2])), -a2 + 
           I (\[Pi] (Conjugate[
                   r1] ((-I + Sqrt[3]/2)^2 + Cos[(2 \[Pi])/9]) + 
                 Conjugate[
                   r2] ((I + Sqrt[3]/2)^2 + Cos[(2 \[Pi])/9])) + 
              3 \[Pi] (Conjugate[
                   r1] ((-2 I + Sqrt[3])^2 + 2 Cos[(2 \[Pi])/9]) + 
                 Conjugate[
                   r2] ((2 I + Sqrt[3])^2 + 2 Cos[(2 \[Pi])/9]))), 
          C2 (1 - n1) - C0 (1 + n1) - 
           4 Im[(A a1 + A1 a2) r1], -C0 (1 + n2) - 
           4 Im[(a2 B + a1 B1) r2], -r1 + 
           I n1 Conjugate[A a1 + a2 B1], -r2 + 
           I n2 Conjugate[a1 A1 + a2 B]} /. {r1 -> r1re + I*r1im, 
          r2 -> r2re + I*r2im, a1 -> a1re + I*a1im, 
          a2 -> a2re + I*a2im}]]] /. 0 -> Nothing]]]
vars = {a1re, a1im, a2re, a2im, r1re, r1im, r2re, r2im, n1, n2};

From here it is straightforward to define a solver that will give numeric output when all parameters are fed explicit numeric values.

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  • $\begingroup$ Thanks Daniel, I will look on to it. $\endgroup$
    – Rupesh
    Commented May 5, 2019 at 13:36

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