The fact that Mathcad did not reduce its result to something simple doesn't mean that there isn't a simple solution.
Had you made the substitution ZL -> RL + j XL in post #3
before you solved for w, you would have had the same problem of a very large output from Mathcad..
But, it isn't necessary to go to all that trouble anyway. When you have an expression such as:
and the zeroes of the part in parentheses are already known, we can see by inspection that multiplying by ZL simply means that ZL=0 is also a solution; the zeroes of the part in parentheses remain the same. Obviously if ZL is a short, you get no output. That's not a solution the original problem wants anyway.
The solution you obtained in post #3 is still a solution, and Zs and ZL don't appear in it.
Even if Mathcad were unable to simplify the very large output, if we know the solution by means of some other method, we can verify that it's a solution by substitution.
I use Mathematica instead of Mathcad, but with either we can verify the solution by substitution:
If these substitutions were made in the very large output you got in post #10, that output would evaluate to zero (assuming no mistakes in the algebra).