Imaginary part of complex number (first post)

bfed · Mar 31, 2010

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

dx · Mar 31, 2010

Use Euler's formula e^iθ = cos(θ) + i sin(θ), and then simplify the resulting expression. The coefficient of i will be the imaginary part of C.

bfed · Mar 31, 2010

so then the imaginary party would be sin(wt)?

and what happens to the A in the function?

dx · Mar 31, 2010

No, thats not right. What is the expression you got after using Euler's formula to expand C?

bfed · Mar 31, 2010

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?

dx · Mar 31, 2010

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.

bfed · Mar 31, 2010

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

dx · Mar 31, 2010

Yep, thats right.

bfed · Mar 31, 2010

good stuff much appreciated dx