Special Relativity Question: Doppler shift Imagine an observer watching a moving rocket carrying on it both, a light source and a clock. If on the rocket the clock is synchronized with the frequency of the light being emitted then will not a stationary observer see not only the clock on the ship slow but then necessarily the wave length of the light increase. Again, imagine those on the rocket are measuring the distance of one wavelength comparing it with some high precision ruler, then will not the observer watch the ruler contract and then necessarily the wavelength of the light along with it, a contradiction to the original conclusion?
 A: 
Again, imagine those on the rocket are measuring the distance of one wavelength comparing it with some high precision ruler, then will not the observer watch the ruler contract and then necessarily the wavelength of the light along with it, a contradiction to the original conclusion?

No. This is an example of the kind of confusion that can result if one imagines that the Lorentz transformation amounts to nothing more than a combination of length contraction and time dilation; if it did, then it would be simply a change of the units of measurement, with no observable significance.
Length contraction by a factor of $\gamma$ is a consequence of the Lorentz transformation, and it is a contraction relative to the length of the object as measured in the object's rest frame. There is no rest frame for a light wave.
A: 
Imagine an observer watching a moving rocket carrying on it both, a light source and a clock. If on the rocket the clock is synchronized with the frequency of the light being emitted

So we suppose not just any clock being carried on the rocket, but a good clock which is characterized (properly) by some particular constant frequency $f_r$; as is the light source carried on the rocket.

then will not a stationary observer [...]

Perhaps we should also assume that the rocket and the "stationary observer" are moving apart from each other at mutually equal speed $v \gt 0$.
Based on the setup described so far, the duration of the "stationary observer" from receiving on signal "tick" from the rocket clock until receiving the next signal "tick" from the rocket clock is 
$$\Delta \tau_s^r := \sqrt{\frac{1 + v/c}{1 - v/c}}~\frac{1}{f_r}.$$

[...] see not only the clock on the ship slow 

"seeing slow" in comparison to what?? This should be specified as well.
Should we also assume that the "stationary observer" carried a clock (and, again, not just any clock, but a good clock) characterized (properly) by some particular constant frequency $f_s$ ?
If so, do you mean by "seeing slow" the comparison 
$$ f_r \lt f_s$$
?
Or do you mean the comparison
$$ f_r \lt \frac{1}{\Delta \tau_s^r} = \sqrt{\frac{1 - v/c}{1 + v/c}}~f_r $$
?
Either way, from the setup described so far, with the "stationary observer", too, carrying a good clock (of frequency $f_s$), it follows that the duration of the rocket from receiving on signal "tick" from the "stationary" clock until receiving the next signal "tick" from the "stationary" clock is 
$$\Delta \tau_r^s := \sqrt{\frac{1 + v/c}{1 - v/c}}~\frac{1}{f_s}.$$

but then necessarily the wave length of the light increase.

Again: "increase" in comparison to what?? ...
