0
$\begingroup$

How can one simplify the following

 Simplify[Conjugate[Sqrt[a^2 - b^2]], {a < b}]

Edit: I want to add that in my case $a$ and $b$ are real and I expect Mathematica to return $-i\sqrt{b^2-a^2}$.

$\endgroup$
5
  • 1
    $\begingroup$ What simplification do you expect? $ \left\{\begin{array}{ll}-i\sqrt{b^2-a^2}& \text{if } |a|<|b|,\\ \sqrt{a^2-b^2} & \text{else}\end{array}\right.$? $\endgroup$
    – NG98
    Jan 27, 2021 at 12:35
  • $\begingroup$ Try FullSimplify[ComplexExpand[Conjugate[Sqrt[a^2 - b^2]], TargetFunctions -> {Re, Im}], a < b] and report back. $\endgroup$ Jan 27, 2021 at 12:42
  • $\begingroup$ I expect Mathematica to return $i\sqrt{b^2-a^2}$given $a$ and $b$ are real. $\endgroup$
    – Mike
    Jan 27, 2021 at 12:56
  • $\begingroup$ If b==1 and a==0 then $i\ \sqrt{b^2-a^2}=i$ which is different from Conjugate[Sqrt[a^2 - b^2]]==-I $\endgroup$
    – Coolwater
    Jan 27, 2021 at 13:50
  • $\begingroup$ Sorry I meant $-i\sqrt{(b^2 - a^2)}$. $\endgroup$
    – Mike
    Jan 27, 2021 at 13:55

2 Answers 2

3
$\begingroup$

A slight modification of @J.M.'s ennui's comment:

FullSimplify[
    ComplexExpand[Conjugate[Sqrt[a^2-b^2]],TargetFunctions->{Re,Im}],
    -b<a<b
]

-I Sqrt[-a^2 + b^2]

$\endgroup$
1
0
$\begingroup$

Try this:

sl = Solve[a^2 - b^2 == -x^2, b][[2, 1]]

(*  b -> Sqrt[a^2 + x^2]  *)

Simplify[Sqrt[a^2 - b^2] /. sl, x > 0] /. x -> Sqrt[-a^2 + b^2]

(* I Sqrt[-a^2 + b^2]  *)

Have fun!

$\endgroup$

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.