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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 $-i\sqrt{b^2-a^2}$.

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    $\begingroup$ 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. $\endgroup$
    – J. M.'s eventual burnout ♦
    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

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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]

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  • $\begingroup$ Yes, this looks way better than what I had. $\endgroup$
    – J. M.'s eventual burnout ♦
    Jan 27, 2021 at 16:26
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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!

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