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Sinval Santos
01/23/02 1:01pm

Im[1]:=Im[(x-a)/Sqrt[Abs[x-b]]]//Simplify[#, x>0&&b>0&&a>0]&

Out[1]=(-a+x) Im[1/ Sqrt[Abs[b-x]]]

a,b,x are real. The difference x-b is also real.

Abs[x-b] it is real and positive.

Therefore: (1 / Sqrt[Abs[b-x]]) it is real and positive.
Therefore the image is zero

Because it doesn't return:
Im[(x-a)/Sqrt[Abs[x-b]]] = 0 ?

How to attribute the 'x' the condition of real number?
Simplify[#,x=Real] & it doesn't work.

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Author Date Posted
Real Number Sinval Santos 01/23/02 1:01pm
Re: Real Number Forum Modera... 01/26/02 3:12pm
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