How to show that imaginary part of complex power is reactive power? Suppose I have a voltage across an element $V(t) = A \cos(\omega t)$ 
and the current through it given by $I(t) = B \cos(\omega t + \phi)$.
The instantaneous power is
$$P(t) = V(t)I(t) = AB \cos(\omega t)\cos(\omega t + \phi).$$
A simple trigonometric identity reduces this to 
$$P(t) = \frac{AB}{2}(\cos(\phi) + \cos(2\omega t + \phi)).$$
The first term is just the average power dissipated in the element and the second term is the power that goes back and forth in the element, 
which as per my understanding, is the reactive power.
Now in the phasor form the power is defined as 
$$\frac{1}{2}\tilde{V}\tilde{I}^* = \frac{AB}{2} e^{-j\phi} =  \frac{AB}{2}(\cos(\phi) - j \sin(\phi)).$$
Clearly the real part here gives the average power dissipated.
But I do not understand why the imaginary part equals the reactive power.
 A: To enlarge slightly upon ad2004's excellent comment, the total power at any instant is related to the sum of the resistive and reactive load responses which plot perpendicularly to one another in the complex plane (in which the phasor representing the total is rotating with frequency omega).
Another way to see this is to realize that the reactive component of the load is responsible for part of the power entering the system from the source to be reflected off the load back to the source, and since this reflected power never makes it into the load, it cannot contribute to the real (resistive) power dissipated in the load.
Let me know if this is not helpful and I'll edit or delete as needed.
A: I believe that some of the discrepancy is due to the fact that the instantaneous power equation that you obtained in the first part of your question needs to be time averaged over one period (i.e. integrated over one period and divided by the period).  This will yield the real power (i.e. that which is actually dissipated in the load) I believe.  I hope this helps. 
