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Finding real part of an expression.

by MathematicalPhysicist
Tags: expression, real
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MathematicalPhysicist
#1
Dec9-12, 07:30 AM
P: 3,220
Hi, I have the next expression:

[tex] \frac{iw+3}{(iw-3)(iw+6)(iw+1)}[/tex]

Now I want to find the real part of this expression via mathematica or maple, and for the love of god it doesn't work, what have I done wrong here?!

the codes and their errors are in the attachments.
http://oi50.tinypic.com/25zs684.jpg
http://oi50.tinypic.com/24wfqt3.jpg

Peace out!

N.B
w is real parameter.
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Bill Simpson
#2
Dec9-12, 03:45 PM
P: 1,030
In[1]:= FullSimplify[Reduce[a+I b==(I w+3)/((I w-3)(I w+6)(I w+1))&& a∈Reals&& b∈Reals&& w∈Reals,{a,b},Backsubstitution->True]]

Out[1]= w∈Reals &&
a == -((54 + 27*w^2 + w^4)/(324 + 369*w^2 + 46*w^4 + w^6)) &&
b == -((w*(-27 + w^2))/(324 + 369*w^2 + 46*w^4 + w^6))

Check this result very carefully before you depend on it
MathematicalPhysicist
#3
Dec9-12, 09:56 PM
P: 3,220
Thanks. Who knew that such a simple task should have a long line of code?!

Hepth
#4
Dec11-12, 05:02 PM
PF Gold
Hepth's Avatar
P: 466
Finding real part of an expression.

What about "ComplexExpand"? ((Mathematica))

ComplexExpand[(I w + 3)/((I w - 3) (I w + 6) (I w + 1))]

[tex]
-\frac{27 w^2}{\left(w^2+1\right) \left(w^2+9\right) \left(w^2+36\right)}-\frac{54}{\left(w^2+1\right) \left(w^2+9\right) \left(w^2+36\right)}-\frac{w^4}{\left(w^2+1\right) \left(w^2+9\right) \left(w^2+36\right)}\\+i \left(\frac{27 w}{\left(w^2+1\right) \left(w^2+9\right) \left(w^2+36\right)}-\frac{w^3}{\left(w^2+1\right) \left(w^2+9\right) \left(w^2+36\right)}\right)
[/tex]


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