# System cannot be solved with methods available

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?

• This is a transcendental equation.I doubt whether it will find exliptic solution. Oct 6, 2015 at 15:13
• You can try plotting the equation and find an approximate solution with FindRoot. Oct 6, 2015 at 15:13
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• 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). Oct 6, 2015 at 20:08

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}]


{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}]


{{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.

• @belisarius.Expand your question? Oct 6, 2015 at 16:12
• If you evaluate Plot[Evaluate[Abs[Subtract @@@ eq]], {P, 0, 100}] The zeroes should be there ... Oct 6, 2015 at 16:18

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