The analysis on page 75 of the book says it all, I believe.

In the differential oscillator case, a full solution (AM + PM) is assumed. This is figuratively an assumption because this is the general solution anyway.
In the Colpitts oscillator case, and on p.75, equation 28 assumes a PM solution only for the gate-source voltage of the transistor. However, equation 29 assumes a general solution for the current in the transistor and the rest of the analysis goes on to validate that AM is very little. Now since the voltage on C1 is a filtered verion of the output voltage then its AM is also very small which validates our initial assumption.

So the analysis flows as follows:
1. Assume no AM on Vc1
2. Assume unknown AM on current pulse
3. Show current pulse AM is small
4. Vc1 AM is filtered version so also small

Note that assuming an AM component on Vc1  (equation 28) will not change the analysis at all. The reason is that equation 29 which calculates the current in the device already assumes an AM component. Therefore, you may modify equation 28 to the following:
Vc1 = V. exp(jwt) + a exp(jw_+ t) - a* exp(jw_- t) + b exp(jw_+ t) + b* exp(jw_- t)    (28_mod)

Now the current waveform is merely the voltage given above by equation (28_mod) multiplied by the transconductance of the device which yields equation (29).

So why did we choose the write equation (28) the way it is? It cleans up the analysis to allow readers to follow.  In fact, on p.78 we hint at this little inconsistency since we deduce a value for the component 'b' in equation (47) which we ignored in Vc1. However, the analysis will not change.

Cheers