How do I use Replace All?

Asked Dec 29 '15 at 18:06

Active Dec 29 '15 at 18:22

Viewed 65 times

I have an expression:

a + b 2^(1/2) + (c + d 2^(1/2))*I

I want to replace each occurrence of 2^(1/2) with -2^(1/2) and each occurrence of I with -I. So I input:

a + b 2^(1/2) + (c + d 2^(1/2)) I /. {2^(1/2) -> -2^(1/2), I -> (-I)}

Mathematica returns:

(* a - Sqrt[2] b - I (c - Sqrt[2] d) *)

as I expected.

Now I want to make the same replacement again on this expression. So I input:

a + b 2^(1/2) + (c + d 2^(1/2)) I /. {2^(1/2) -> -2^(1/2), 
  I -> (-I)}/. {2^(1/2) -> -2^(1/2), I -> (-I)}

Mathematica returns:

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

I do not expect the MINUS sign in the middle of this expression. It seems like it should be PLUS.?

edited Dec 29 '15 at 18:22

bbgodfrey

asked Dec 29 '15 at 18:06

Geoffrey Critzer

1

Please format your code. Edit your post and apply the {} button where needed. – Edmund Dec 29 '15 at 18:11
Perhaps a + b 2^(1/2) + (c + d 2^(1/2)) I /. {2^(1/2) -> (-2)^(1/2), I -> (-I)} – Dr. belisarius Dec 29 '15 at 18:15
I want to replace 2^(1/2) with -2^(1/2). – Geoffrey Critzer Dec 29 '15 at 18:17
Here is a better way to pose my question. If I input I /. I -> -I /. I -> -I I would expect Mathematica to replace I with -I two times and then return I (positive) ? – Geoffrey Critzer Dec 29 '15 at 18:21
2

This is a common problem with replacing complex numbers, and it is has been answered before. In the second replacement, try Complex[0, -1] -> I. – march Dec 29 '15 at 18:24
5

See this or this other or this third – Daniel Lichtblau Dec 29 '15 at 18:36
1

more generally c_Complex:> Conjugate[c] ( which is in Daniels third link.. ) – george2079 Dec 29 '15 at 19:10

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