Complex Notation Homework: Solve for B & Phi in Terms of A, Omega, Delta

[w3390]

Homework Statement

If x= Acos(\omegat + \delta), then one can also write it as x = Re(Be^{i\Phi}). Find B and \Phi in terms of A, \omega, and \delta if B is real.

Homework Equations

The Attempt at a Solution

Not sure where to start on this one. I know you guys can't give answers. All I'm looking for is where to get started. Any help would be appreciated.

 

[berkeman]

Homework Statement

If x= Acos(\omegat + \delta), then one can also write it as x = Re(Be^{i\Phi}). Find B and \Phi in terms of A, \omega, and \delta if B is real.

Homework Equations

The Attempt at a Solution

Not sure where to start on this one. I know you guys can't give answers. All I'm looking for is where to get started. Any help would be appreciated.

Are you familiar with converting between the rectangular and polar forms of complex numbers? See partway down this wiki page:

http://en.wikipedia.org/wiki/Polar_coordinate_system

.

 

[w3390]

Actually, I think I might have something.

Euler's formula says: e^(i*phi) = cos(phi) + i*sin(phi)

The real part of this is: Re(e^(i*phi)) = cos(phi).

Therefore, the real part of Be^(i*phi) is: Bcos(phi).

So I have: X = Bcos(phi) and X = Acos(wt + delta)

Am I able to just compare the two equations to get the following relationships:

B = A

PHI = wt + deltaIt can't be that simple, can it?

 

[berkeman]

Actually, I think I might have something.

Euler's formula says: e^(i*phi) = cos(phi) + i*sin(phi)

The real part of this is: Re(e^(i*phi)) = cos(phi).

Therefore, the real part of Be^(i*phi) is: Bcos(phi).

So I have: X = Bcos(phi) and X = Acos(wt + delta)

Am I able to just compare the two equations to get the following relationships:

B = A

PHI = wt + delta


It can't be that simple, can it?