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I'm trying to solve the following equation:

eqn = (-(Sqrt[3 - P] - I*Sqrt[P]) + (Sqrt[3 - P] + I*Sqrt[P])*
  Exp[-2*Pi*Sqrt[3 - P]])/((Sqrt[3 - P] - 
   I*Sqrt[P]) + (Sqrt[3 - P] + I*Sqrt[P])*
  Exp[-2*Pi*Sqrt[3 - P]]) == I*Sqrt[P/(3 - P)]*(1 + Exp[-2*I*Pi*Sqrt[P]])/(1 - 
  Exp[-2*I*Pi*Sqrt[P]]);

eqn //TeXForm

$\frac{e^{-2 \pi \sqrt{3-P}} \left(\sqrt{3-P}+i \sqrt{P}\right)-\sqrt{3-P}+i \sqrt{P}}{e^{-2 \pi \sqrt{3-P}} \left(\sqrt{3-P}+i \sqrt{P}\right)+\sqrt{3-P}-i \sqrt{P}}=\frac{i \left(1+e^{-2 i \pi \sqrt{P}}\right) \sqrt{\frac{P}{3-P}}}{1-e^{-2 i \pi \sqrt{P}}}$

The command Solve hasn't been successful so far:

Solve[eqn, P]

Any ideas?

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  • 1
    $\begingroup$ This is a transcendental equation.I doubt whether it will find exliptic solution. $\endgroup$
    – Mariusz Iwaniuk
    Oct 6, 2015 at 15:13
  • 2
    $\begingroup$ You can try plotting the equation and find an approximate solution with FindRoot. $\endgroup$
    – m0nhawk
    Oct 6, 2015 at 15:13
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 0) Browse the common pitfalls question 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 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! $\endgroup$
    – Dr. belisarius
    Oct 6, 2015 at 15:47
  • $\begingroup$ It helps to start by visualising your function (difference between the left and right sides of your equation) by creating contour plots in the complex P plane, so you can see where the function's zeros are likely to be. For instance, you could start with ContourPlot[Abs[func[x + I y]], {x, 0, 10}, {y, -2, 2}] (Arg is done analogously). $\endgroup$
    – Stephen Luttrell
    Oct 6, 2015 at 20:08

2 Answers 2

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Solve, Reduce, NSolve can not solve this problem; this is a transcendental equation. It can only be solved numerically.

$$\left\{\frac{e^{-2 \pi \sqrt{3-P}} \left(\sqrt{3-P}+i \sqrt{P}\right)-\sqrt{3-P}+i \sqrt{P}}{e^{-2 \pi \sqrt{3-P}} \left(\sqrt{3-P}+i \sqrt{P}\right)+\sqrt{3-P}-i \sqrt{P}}-\frac{i \left(1+e^{-2 i \pi \sqrt{P}}\right) \sqrt{\frac{P}{3-P}}}{1-e^{-2 i \pi \sqrt{P}}}=0\right\}$$

 eq = {(-(Sqrt[3 - P] - I*Sqrt[P]) + (Sqrt[3 - P] + I*Sqrt[P])*
    Exp[-2*Pi*Sqrt[3 - P]])/((Sqrt[3 - P] - 
     I*Sqrt[P]) + (Sqrt[3 - P] + I*Sqrt[P])*
    Exp[-2*Pi*Sqrt[3 - P]]) - 
 I Sqrt[P/(3 - P)] (1 + Exp[-2*I Pi Sqrt[P]])/(1 - 
    Exp[-2*I Pi Sqrt[P]]) == 0}

In the real domain:

  Plot[Evaluate[Re@eq[[1, 1]]], {P, 0, 4}]

enter image description here

{FindRoot[Re@eq[[1, 1]], {P, 0.5}], FindRoot[Re@eq[[1, 1]], {P, 2.5}]}

{{P -> 0.703961}, {P -> 2.62104}}

In the complex domain:

 Plot[Evaluate[Im@eq[[1, 1]]], {P, 5, 30}]

enter image description here

{{P -> 6.54838}, {P -> 12.1886}, {P -> 20.0831}}

As you can see from this plot, the equation has a infinitely many solutions in the complex domain.

enter image description here

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  • $\begingroup$ @belisarius.Expand your question? $\endgroup$
    – Mariusz Iwaniuk
    Oct 6, 2015 at 16:12
  • $\begingroup$ If you evaluate Plot[Evaluate[Abs[Subtract @@@ eq]], {P, 0, 100}] The zeroes should be there ... $\endgroup$
    – Dr. belisarius
    Oct 6, 2015 at 16:18
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You need to include a domain restriction for Solve to find any solutions:

zeros = P /. Solve[eq && -5 < Re[P] < 5 && -5 < Im[P] < 5, P];

Solve::fexp: Warning: Solve used FunctionExpand to transform the system. Since FunctionExpand transformation rules are only generically correct, the solution set might have been altered.

Solve::incs: Warning: Solve was unable to prove that the solution set found is complete.

The result zeros is a complicated looking set of Root objects which can be numericized to any desired precision. At machine precision we have:

N[zeros]

{0.703961 - 0.0000232491 I, 2.62093 - 0.00664407 I, 3.95346 - 0.275317 I}

Let's check numerically if zeros satisfies the equation:

eq /. P -> N[zeros, 100]

True

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