RE: how mathematica deals with complex i in output
- To: mathgroup at smc.vnet.net
- Subject: [mg34346] RE: [mg34334] how mathematica deals with complex i in output
- From: "Florian Jaccard" <jaccardf at eicn.ch>
- Date: Thu, 16 May 2002 05:08:28 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Try this :
y = ((I/4)*p^2*Log[1 - (I*Sqrt[a]*(p + qD))/Sqrt[-(a*p^2) -
\[Omega]]])/(a^(3/2)*Sqrt[-(a*p^2) - \[Omega]])
Simplify[y, {qD > 0, a > 0, p > 0, \[Omega] > 0}]
You have to be sure that the assumptions are allowed... If it is the case,
it work's !
Meilleures salutations
Florian Jaccard
e-mail : jaccardf at eicn.ch
-----Message d'origine-----
De : Nicolas Bock [mailto:nbock at acsu.buffalo.edu]
Envoyé : mer., 15. mai 2002 09:35
À : mathgroup at smc.vnet.net
Objet : [mg34334] how mathematica deals with complex i in output
Hello,
After integrating an expression, mathematica gave me the following:
(Local) Out[5]//InputForm=
((I/4)*p^2*Log[1 - (I*Sqrt[a]*(p + qD))/Sqrt[-(a*p^2) -
\[Omega]]])/(a^(3/2)*Sqrt[-(a*p^2) - \[Omega]])
Sorry about the somewhat confusing form of this expression, but I
don't know how to do this better in an email. Anyway, my question is
this: There are a number of I's in the output, which would make you
believe that the output is something complex. Upon closer inspection
though one notices that it is actually not, since all the factors of I
cancel from this expression, if the definition of I is used, I ==
sqrt(-1). How can I tell mathematica to do the same and to eliminate
all those factors of I for me? I already tried a number of things, for
example RealOnly, ImRe, but so far nothing worked. I guess this is a
problem of how mathematica formats its output and how it arranges the
terms in the output, but I don't know how to change that behavior.
Thanks already for any suggestions, nick