0
expr=(y+Im[-2 (α + I β) -
    Sqrt[2] (α + I β)  Sech[
      Sqrt[2] (α + I β)]^2 -
    Tanh[Sqrt[2] (α + I β)]])/. Solve[Re[-2 (α + I β) -
    Sqrt[2] (α + I β)  Sech[
      Sqrt[2] (α + I β)]^2 -
    Tanh[Sqrt[2] (α + I β)]]==0,{α,Reals}]
Solve[expr==0,{β,Reals}]

I could not make Mathematica compute both $\alpha$ and $\beta$ in terms of $y.$ Besides , Solve command; I also tried to make it work through Reduce, and Solve always. If you were to complex expand after trigExpand that expression we have in $Re$ and $Im$ then, we have the following associated equation that I would use the solution of it to get $\beta$ in terms of $\alpha:$

Reduce[-α - ( 4 α Cos[Sqrt[2] β]^2 Cosh[
 Sqrt[2] α]^2)/(Cos[2 Sqrt[2] β] +  Cosh[2 Sqrt[2] α])^2 + ( 4 Sqrt[2] α Cos[Sqrt[2] β]^2 Cosh[
 Sqrt[2] α]^2)/(Cos[2 Sqrt[2] β] +  Cosh[2 Sqrt[2] α])^2 + (α Sin[2 Sqrt[2] β]^2)/(Cos[2 Sqrt[2] β] + 
 Cosh[2 Sqrt[2] α])^2 - (8 β Cos[Sqrt[2] β] Cosh[Sqrt[2] α] Sin[ Sqrt[2] β] Sinh[ Sqrt[2] α])/(Cos[2 Sqrt[2] [Beta]] +   Cosh[2 Sqrt[2] α])^2 + ( 8 Sqrt[2] β Cos[Sqrt[2] β] Cosh[Sqrt[2] α] Sin[ Sqrt[2] β] Sinh[ Sqrt[2] α])/(Cos[2 Sqrt[2] β] +  Cosh[2 Sqrt[2] α])^2 + (4 α Sin[Sqrt[2] β]^2 Sinh[ Sqrt[2] α]^2)/(Cos[2 Sqrt[2] β] + 
 Cosh[2 Sqrt[2] α])^2 - ( 4 Sqrt[2] α Sin[Sqrt[2] β]^2 Sinh[ Sqrt[2] α]^2)/(Cos[2 Sqrt[2] β] + Cosh[2 Sqrt[2] [Alpha]])^2 - Sinh[2 Sqrt[2] α]/(Cos[2 Sqrt[2] β] + Cosh[2 Sqrt[2] α]) + ( 2 β Sin[2 Sqrt[2] β] Sinh[ 2 Sqrt[2] α])/(Cos[2 Sqrt[2] β] +  Cosh[2 Sqrt[2] α])^2 - (α Sinh[ 2 Sqrt[2] α]^2)/(Cos[2 Sqrt[2] β] +  Cosh[2 Sqrt[2] α])^2 == 0, α, Reals]

I also tried some other ways in order to visualize how the intersection of those contours change based on different values of $y.$ However, this does not make to much sense for my main goal.

LCarvalho
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user92604
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  • Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – Michael E2 Aug 09 '16 at 19:20
  • For related aspects see this answer: Solve symbolically a transcendental trigonometric equation and plot its solutions. By the way your question is not well posed. Are α and β real, are they bounded? The result of Solve are replacement rules so what do you expect by α -> Solve[...]? – Artes Aug 10 '16 at 00:01
  • @Artes,Thanks for telling me that I should have told where $\alpha,\beta$ lie in. However, I already checked out all the questions except for the recent ones regarding this problem. In that question it is really easy to tell Mathematica what to solve , because there is only only one trig function and a linear term depending on another variable. As I explained within the question, I figured out how y changes the zeroes for the real and imaginary parts numerically and on plots. I have hard time with getting the symbolic relats between $\alpha, \beta$ and $y.$ I can attach plots, if you want. – user92604 Aug 10 '16 at 01:02

0 Answers0