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Imaginary part of complex number (first post)

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data
    C=A*e^(-i*wt)*sin(k*x); A,w,t,k,x are real numbers. Find imaginary part.


    2. Relevant equations



    3. The attempt at a solution
    Im(C)=cos(wt)-i*sin(wt)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 31, 2010 #2

    dx

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    Use Euler's formula e = cos(θ) + i sin(θ), and then simplify the resulting expression. The coefficient of i will be the imaginary part of C.
     
  4. Mar 31, 2010 #3
    so then the imaginary party would be sin(wt)?

    and what happens to the A in the function?
     
  5. Mar 31, 2010 #4

    dx

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    No, thats not right. What is the expression you got after using Euler's formula to expand C?
     
  6. Mar 31, 2010 #5
    e^(-iwt)=cos(wt)-i*sin(wt) is how I think to the Euler formula...

    then do i substitute it back into the original expression?
     
  7. Mar 31, 2010 #6

    dx

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    Yes, substitue it back into the original expression, and then expand out the brackets using the distributive law of multiplication, i.e. A(B + C) = AB + AC.

    Then you will have an expression of the form C = R + iI, and I is the imaginary part of C.
     
  8. Mar 31, 2010 #7
    so...

    C=A*cos(wt)*sin(kx)-i*A*sin(wt)*sin(kx)

    Re(C)=A*cos(wt)*sin(kx)
    and
    Im(C)=-A*sin(wt)*sin(kx)

    is this the proper solution?

    And thanks dx
     
  9. Mar 31, 2010 #8

    dx

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    Yep, thats right.
     
  10. Mar 31, 2010 #9
    good stuff much appreciated dx
     
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