1
$\begingroup$

I think it is straightforward from the title,

Simplify[Im[1/(x+I)], x > 0]

spits out,

Im[1/(x+I)]

while I would have liked,

- 1/(x^2+1)

Isn't there any easy way to get this result? I think there should be something simpler than dividing by the complex conjugate of the denominator, simplifying, then multiplying by the conjugate of the denominator and finally simplifying again.

$\endgroup$
2
$\begingroup$

Use ComplexExpand to simplify, assuming variables are real

ComplexExpand[Im[1/(x + I)]]
(* -(1/(1 + x^2)) *)
| improve this answer | |
$\endgroup$
  • $\begingroup$ Nice, that seems to work well, but that's a shame Simplify or Refine don't do it. $\endgroup$ – Olivier Massicot Sep 10 '19 at 0:20
  • 1
    $\begingroup$ @OlivierMassicot - Evaluate Greater @@ (LeafCount /@ {ComplexExpand[Im[1/(x + I)]], Simplify[Im[1/(x + I)], x > 0]}) and you will see why. ComplexExpand gives you easier control over the form of complex expressions, particularly if you use the option TargetFunctions. $\endgroup$ – Bob Hanlon Sep 10 '19 at 5:30

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.