The result will indeed be a high pass filter. From your difference equation, the transfer function of the low pass filter is
Note that this is actually a leaky integrator, not a classic low pass filter, because its frequency response does not have a zero at Nyquist.
The high pass filter has the transfer function
which has a zero at DC (i.e. at $z=1$), as it should. However, it is scaled, i.e. its value at Nyquist ($z=-1$) is not $1$, but $2(1-\beta)/(2-\beta)$. So in order to have a 0 dB gain at Nyquist, you need to scale your high pass filter with the inverse of that factor:
You can compute the 3dB cut-off frequencies of the low pass filter (1) and of the high pass filter (2) by solving
After some algebra this gives for the low pass filter
and for the high pass filter
where $\omega$ is the normalized frequency in radians:
with the sampling frequency $f_s$.