Here is what I have used for this.
From this link, Why do I get empty graph when adding 'Abs' function? I know that I can remove Abs with ComplexExpand

ComplexExpand[Sqrt[1 - Abs[x]^2]]


((1 - x^2)^2)^(1/4) Cos[1/2 Arg[1 - Abs[x]^2]] + 
 I ((1 - x^2)^2)^(1/4) Sin[1/2 Arg[1 - Abs[x]^2]]

However, it works just fine when I use Plus in the Sqrt.

ComplexExpand[Sqrt[1 + Abs[x]^2]]


Sqrt[1 + x^2]

Is there anyone who can explain this?

  • $\begingroup$ Maybe you can try setting the option TargetFunctions->{Re, Im} to ComplexExpand? $\endgroup$ – QuantumDot Jul 26 at 5:44
  • 1
    $\begingroup$ Documentation says ComplexExpand assumes all variables are real. Since your example doesn't work, perhaps they missed a case. Or you can use Simplify[Sqrt[1+ Abs[x]^2],Element[x,Reals]] which works with both + and - and get on with the work you need to do. $\endgroup$ – Bill Jul 26 at 6:02
  • $\begingroup$ @QuantumDot, your method is not working $\endgroup$ – kile Jul 26 at 6:16

Maybe adding Simplify gives you what you want.

ComplexExpand[Sqrt[1 - Abs[x]^2]] // Simplify
(*Piecewise[{{I*((x^2 - 1)^2)^(1/4), Abs[x] > 1}}, ((x^2 - 1)^2)^(1/4)]*)


$Assumptions = -1 < x < 1

ComplexExpand[Sqrt[1 - Abs[x]^2]] // Simplify
(*Sqrt[1 - x^2]*)

The plus case is different because the value inside the Sqrt is always positive. It is generally a good idea to add Simplify to any result ComplexExpand returns.

| improve this answer | |
  • $\begingroup$ I think there is no need to add ComplexExpand. You pan get the same result with only Simplify used here $\endgroup$ – kile Jul 27 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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