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Finding real part of an expression.

  1. Dec 9, 2012 #1


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    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.

    Peace out!

    w is real parameter.
  2. jcsd
  3. Dec 9, 2012 #2
    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
    Last edited: Dec 9, 2012
  4. Dec 9, 2012 #3


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    Thanks. Who knew that such a simple task should have a long line of code?!
  5. Dec 11, 2012 #4


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    What about "ComplexExpand"? ((Mathematica))

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

    -\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)
    Last edited: Dec 11, 2012
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