# Finding real part of an expression.

by MathematicalPhysicist
Tags: expression, real
 P: 3,128 Hi, I have the next expression: $$\frac{iw+3}{(iw-3)(iw+6)(iw+1)}$$ 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.
 P: 935 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
 P: 3,128 Thanks. Who knew that such a simple task should have a long line of code?!
PF Patron
P: 413

## Finding real part of an expression.

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