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

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

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

## 2 Answers

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]

• Yes, this looks way better than what I had. Jan 27, 2021 at 16:26

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!