I have following set of coupled transcendental equations.

$\tan^{-1} \left(\frac{\sigma+\beta}{K'-1}\right)+\tan^{-1} \left(\frac{\sigma-\beta}{K'-1}\right)=\frac{2 \sigma}{K'}\cos(\beta\tau)$
$\log\left(\frac{(K'-1)^2+(\sigma-\beta)^2}{(K'-1)^2+(\sigma+\beta)^2}\right)=\frac{4 \sigma}{K'}\sin(\beta\tau)$

Here K', $\sigma$ and $\beta$ are variables and $\tau=0.07$ is a constant. I want to obtain $\sigma$ vs K' curve (as shown in the attached figure) by eliminating $\beta$.

I tried to eliminate $\beta$ by using Eliminate function of Mathematica, but it gives:

Eliminate::ifun: Inverse functions are being used by Eliminate, so some solutions may not be found; use Reduce for complete solution information.

When a tried using Reduce, Mathematica keeps on running and does not give anything.

I also tried using FindRoot and NSolve to find values of K' and $\sigma$ that satisfy these equations for a range of $\beta$ going from $-2\pi/\tau$ to $2\pi/\tau$, but I keep on get either $\sigma\approx 0$ or very high.

Do[Print[FindRoot[{ArcTan[(sigma + beta)/(k - 1)] + 
  ArcTan[(sigma - beta)/(k - 1)] == ((2 sigma) Cos[0.07  beta])/k,
 Log[((k - 1)^2 + (sigma - beta)^2)/((k - 1)^2 + (sigma + 
     beta)^2)] == ((4 sigma) Sin[0.07  beta])/k}, {{sigma, 
 2.5}, {k, 2.0}}]], {beta, -2 Pi/0.07, 2 Pi/0.07, 0.01}]

I am relatively new to Mathematica, so please suggest me ways to handle such equations.enter image description here

  • 2
    $\begingroup$ Please provide Mathematica code with/instead of latex equations. $\endgroup$
    – Öskå
    Commented Jun 3, 2014 at 18:52
  • $\begingroup$ I don't have Mathematica at hand right now, but you may try to divide your equations by sigma (see my answer below), so that sigma=0 is no longer a solution to your system. $\endgroup$
    – user10957
    Commented Jun 4, 2014 at 7:41
  • $\begingroup$ @Jean-ClaudeArbaut Thank you very much for your answer. I used your suggestion of dividing both equations by (1/sigma). Though making surface plots for the functions and looking at their intersection at [0,0] plane didn't help in this case but it really helped for a similar set of equations. For this particular case, the surfaces do not intersect at any point other than sigma equal to zero. Anyways, I can't thank you enough for your help. $\endgroup$
    – miki
    Commented Jun 6, 2014 at 18:22

1 Answer 1


Notice σ=0 is always a solution, so the problem is not well-posed.

It may help to have a look at the two equations separately:

ContourPlot3D[ArcTan[(σ + β)/(k - 1)] + 
  ArcTan[(σ - β)/(k - 1)] == ((2 σ)
  Cos[0.07 β])/k,
  {σ, -5, 5}, {β, -5, 5}, {k, -5, 5}]

enter image description here

ContourPlot3D[Log[((k - 1)^2 + (σ - β)^2)/
  ((k - 1)^2 + (σ + β)^2)] ==
  4 σ Sin[0.07 β]/k,
  {σ, -5, 5}, {β, -5, 5}, {k, -5, 5}]

enter image description here


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