Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to check when the expression

$$ \sqrt{\sqrt{\lambda^{2}\left(2E+\lambda^{2}\right)}+E+\lambda^{2}} $$

is real, when it is purely imaginary and when it is complex (with imaginary part not zero).

Also I would like to know the sign of the imaginary part in the case it is non-zero. To check the last part, after some steps, I am using Reduce for inequalities like $\mathrm{Im}\left(\sqrt{\lambda^{2}\left(2E+\lambda^{2}\right)}\right)>0 $ but this is taking a lot of time on Mathematica. Is there a way to get the kind of results that I want more quickly?

The variables are real and the code I ran is

  Reduce[ Im[ Sqrt[ e + λ^2  -Sqrt[ λ^2 (2 e + λ^2)]]] > 0, {e, λ}]
share|improve this question
Can you include the code which is taking a long time? – Jonathan Shock Jul 16 '13 at 17:11
Are the variables real ? – b.gatessucks Jul 16 '13 at 17:16
Reduce[Im[Sqrt[e + x^2 + Sqrt[(x^2) (2*e + x^2)]]] == 0, x, Reals] for real vars – Dr. belisarius Jul 16 '13 at 17:55
@belisarius Interestingly, Reduce doesn't seem to solve the <0 case correctly. – Sjoerd C. de Vries Jul 16 '13 at 20:29

Let's see what are the regions of interest:

n = 4;
   RegionPlot[#[[2]][#[[1]][Sqrt[e + x^2 + Sqrt[(x^2) (2*e + x^2)]]], 0], 
             {x, -n, n}, {e, -n, n}, PlotLabel -> #, 
             AxesLabel -> Automatic, LabelStyle -> Medium] & /@ 
                                                 Tuples@{{Re, Im}, {Less, Greater}},
   2] // Grid // Framed

Mathematica graphics

So in principle we should only focus on the sign of the imaginary part. Like this:

Reduce[Im[Sqrt[e + x^2 + Sqrt[(x^2) (2*e + x^2)]]] == 0, e, Reals]
 e >= -(x^2/2)

and that's the information you were after.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.