1
$\begingroup$

I am trying to validate this property:

$\sqrt{1-z}=\sqrt{-(-1+z)}=(\text{}\pm i) \sqrt{-1+z}$

FullSimplify[Sqrt[1 - z] == I Sqrt[(-1 + z)]]

I have seen how we need to be specially careful when trying to check equalities among square root expressions, and generally using Assumptions and setting some values as Reals,Positives, etc, gets the job done.

However I can't say for sure what would I need to add as an assumption in this case, since there are no other parameters than the constants i and 1.

Thanks in advance!

$\endgroup$

1 Answer 1

1
$\begingroup$

The Sqrt of a number -- like all functions -- is single-valued for a given input. The root is taken to minimize its phase, so whether it is plus or minus depends on the input.

FullSimplify[Sqrt[1 - z] == I Sqrt[(-1 + z)] /. z -> #] & /@ {5 + 3 I, -5 + 
   3 I, 5 - 3 I, -5 - 3 I}

(* {False, False, True, True} *)

FullSimplify[Sqrt[1 - z] == -I Sqrt[(-1 + z)] /. z -> #] & /@ {5 + 3 I, -5 + 
   3 I, 5 - 3 I, -5 - 3 I}

{True, True, False, False}

Assuming[z > 1, Sqrt[1 - z] == I Sqrt[(-1 + z)] // FullSimplify]

(* True *)

Assuming[z < 1, Sqrt[1 - z] == -I Sqrt[(-1 + z)] // FullSimplify]

(* True *)

Graphically,

ReImPlot[#, {z, -5, 5}] & /@ {Sqrt[1 - z], I Sqrt[(-1 + z)]}

enter image description here

ReImPlot[#, {z, -5, 5}] & /@ {Sqrt[1 - z], -I Sqrt[(-1 + z)]}

enter image description here

$\endgroup$
1
  • $\begingroup$ This is an interesting way of seeing it, thanks for your complete explanation! $\endgroup$
    – Jmtz
    Commented Jul 29, 2023 at 0:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.