Why does this not work: Alt. way to compute the period of a discrete periodic signal that is a sum of 2 complex exponentials I want to compute the fundamental period of the following discrete time signal:

$ x[n]= \exp^{(j\frac{2\pi}{3})n} + \exp^{(j\frac{3\pi}{4})n}$

I know I can do this by taking the fundamental periods of the first term $\exp^{(j\frac{2\pi}{3})n}$, $N_1=3$ and of the second term $\exp^{(j\frac{3\pi}{4})n}$ $N_2=8$ and taking the least commmon multiple which gives $N=24$.

However I tried to express $x[n]$ as a single term as: 

$ x[n]=\exp^{(j\frac{8.5\pi}{12})n} (\exp^{(j\frac{0.5\pi}{12})n}-\exp^{(j\frac{-0.5\pi}{12})n}) $

$ x[n]=2\exp^{(j\frac{8.5\pi}{12})n}.\cos(\frac{0.5 \pi n}{12}) $

Since the modulus of the complex exponential represents the magnitude of the signal, the fundamental period then equates to $N=48$ using only the cosine term. I tried this for varying frequencies and using this method I always got a the value for the period that was twice the correct value.

I know there is a flaw in my reasoning but could someone point that out ? And is there a reason why the fundamental period computed in this way always comes out to be twice that of the correct value ?
The flaw in your reasoning is that you only consider the cosine term. The cosine term has a period of $48$. But there's also the complex exponential, which is not just a magnitude but a function of $n$. You can show that that exponential term is periodic with a period of $48$. However, you cannot conclude that the multiplication of the two terms must also have a period of $48$.

Multiplying the two terms results in a period of $24$ because

$$e^{j\frac{8.5\pi}{12}n}=-e^{j\frac{8.5\pi}{12}(n+24)}$$

and

$$\cos\left(\frac{0.5\pi n}{12}\right)=-\cos\left(\frac{0.5\pi (n+24)}{12}\right)$$

so the sign changes cancel in the product.