# Imaginary part of complex number (first post)

• bfed
In summary, to find the imaginary part of the function C = A*e^(-i*wt)*sin(k*x), use Euler's formula eiθ = cos(θ) + i sin(θ) to expand the expression and then simplify it. The coefficient of i will be the imaginary part of C. After substituting the expanded expression back into the original, use the distributive law of multiplication to expand the brackets. The resulting expression will be of the form C = R + iI, where I is the imaginary part of C.

## Homework Statement

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

## The Attempt at a Solution

Im(C)=cos(wt)-i*sin(wt)

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