I want to get the Real part of this expression - shouldn't be too hard to evaluate. Why is Mathematica not evaluating but returning the same code?

In == Re[χ/(1 + I ω τ)]
Out == Re[χ/(1 + I ω τ)]

Thank you!

  • $\begingroup$ To directly answer the question as asked: Mathematica has no way of knowing that you intend the symbolic entities $\chi$, $\omega$, and $\tau$ to be real; without further information, such entities might be complex. $\endgroup$ – murray Dec 7 '14 at 17:50
  • $\begingroup$ Same as this post $\endgroup$ – SquareOne Dec 7 '14 at 17:58

Look at the documentation for Re, under Possible Issues

Re can stay unevaluated for numeric arguments: {Re[Log[2 + I]], Re[Sqrt[1 + I]]}

To get around this, try using Re[ComplexExpand[χ/(1 + I ω τ)]].

|improve this answer|||||
  • 3
    $\begingroup$ To go one step further, you can display the real part by assuming the variables are real (just like ComplexExpand did): Refine[Re[ComplexExpand[\[Chi]/(1+I \[Omega] \[Tau])]],{\[Chi],\[Omega],\[Tau]}\[Element]Reals] $\endgroup$ – seismatica Dec 7 '14 at 8:02
  • $\begingroup$ Sorry, I accidentally downvoted and didn't notice fast enough. Reversed :-) $\endgroup$ – Yves Klett Dec 7 '14 at 20:06

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.