Determine filter given transfer function
No $$H(z) = 1 + \frac{\alpha}{1-\alpha z^{-1}}$$ is not a lowpass filter. It's a lowpass filter in parallel with an all pass filter (kind of low-boost shelving filter). If you want a lowpass filter which rejects the high frequencies then you may take the following:

$$H_{lp}(z) = \frac{\alpha}{1-\alpha z^{-1}}$$ which is a lowpass filter for the range of parameter $\alpha$. And therefore $1 - H_{lp}(z)$ will be the complementary highpass filter in simplest terms.

So you can see it by the following code:

freqz(alfa, [1 -alfa])


Note that the gain of the filter will be dependent on $\alpha$! Note also that as Matt.L commented you might have $1 + \alpha$ in the numerator instead...