Imaginary part of complex number (first post)

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To find the imaginary part of the complex number C=A*e^(-i*wt)*sin(k*x), Euler's formula is applied, resulting in e^(-iwt)=cos(wt)-i*sin(wt). Substituting this back into the original expression and expanding using the distributive law yields C=A*cos(wt)*sin(kx)-i*A*sin(wt)*sin(kx). The real part is Re(C)=A*cos(wt)*sin(kx), while the imaginary part is Im(C)=-A*sin(wt)*sin(kx). This confirms the proper solution for the imaginary part of C.
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Homework Statement


C=A*e^(-i*wt)*sin(k*x); A,w,t,k,x are real numbers. Find imaginary part.


Homework Equations





The Attempt at a Solution


Im(C)=cos(wt)-i*sin(wt)
 
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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.
 
so then the imaginary party would be sin(wt)?

and what happens to the A in the function?
 
No, that's not right. What is the expression you got after using Euler's formula to expand C?
 
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?
 
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.
 
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
 
Yep, that's right.
 
good stuff much appreciated dx
 

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