Although you are in frequency domain you still should be able to get all parameters as you were in time domain. They are different domains but they both should represent the same thing. Time domain represent things in terms of amplitude in respect to time. Frequency domain represent things in terms of amplitude AND PHASE in respect to frequency values. Note that you should have both amplitude and phase in frequency domain, since in the time domain the phase can be represented in the same plot by a shift.
One way to represent these things in frequency domain is by dealing with complex numbers. Complex numbers can be viewed as vectors in a 2D space which have a length (as you said) and an angle. The length represents the output/input ratio and the angle represent the phase shift in comparison also to the input.
So, answering your question, you should calculate the H length to find your output/input ratio. To help you, imagine that:
\$e^{jw}=cos(w)+jsin(w)\$
In other words, its a complex number with always length of 1 and angle \$w\$
You can solve this by two methods:
-Vector method:
imagine that number 1 is \$Z=1+0.i\$ which is a vector to the right, with length 1 and angle \$0\$.
Imagine that \$e^{jw}\$ is a vector that I showed right above
Now add them. Then divive vectors 1 by the vector that you've found.
-Cartesian Coordinates:
represent all in terms of \$Z=a+jb\$ and also \$e^{jw}=cos(w)+jsin(w)\$
and imagine that you have:
\$\large Z = \frac{Z_1}{Z_2+Z_3}\$
and then find length of Z by:
\$|Z| = \sqrt{a^2+b^2}\$
You don't have enough information to know if the system is LTI or not.
To know a system is linear, you have to know that for any input \$x(n)\$ producing an output \$y(n)\$, then a scaled input \$A\cdot{}x(n)\$ produces the output \$A\cdot{}y(n)\$, no matter what input was chosen for \$x(n)\$. Since you only know the output for one particular input, you don't know if your system is linear or not.
However, the point is largely moot. We normally don't try to prove a system is LTI or not. We just assume a system is (approximately) LTI in order to model its behavior. Real systems are usually nonlinear (because a large enough input will produce saturation or damage the system or cause numerical overflow) and time-variant (because they are turned on and off from time to time). So LTI-ness is just an approximation to the real behavior, or a given of a pedagogical problem, to allow modeling the system.
Best Answer
I'm assuming you are trying to get a bode plot which shows the magnitude and phase response you would get if you were to test the the system by slowly sweeping the frequency a constant amplitude sine wave input and measuring the output amplitude and phase at various points.
If this is the case then there is only one frequency \$ \omega \$. Now we know that \$ e^{j \theta} = \cos(\theta) + i \cdot sin(\theta)\$.
You can therefore treat each exponential term as a vector with amplitude and phase and add them as vectors.
Note however, that if you compare the results with a real digital filter you will not get exactly the same result because the digital filter is only taking measurements at certain instants in time (the sampling frequency) and the data is quantised. However providing the sampling frequency is much larger than the frequency of interest it is a reasonable approximation.