single zero in z-domain transfer function calculatio
The transfer function based on poles and zeroes works just like in the s-plane, with proper adjustments for the z-plane: both the numerator and the denominator is built by using the product formula:
$$H(z)=K\dfrac{\prod_{k=0}^M{1-a[k]z^{-1}}}{\prod_{k=1}^N{1-b[k]z^{-1}}} \tag{1}$$
In your case, since both their values are real, your zero is the value 0.1 (if that's what you want, a zero at one tenth of the sampling frequency). It should be clear that if using 0.533 in the numerator gives you a zero at 0.533, its value is that number.
What you're saying about the Euler's formula is related to evaluating the z in the frequency domain, but your business is with the zero, not z.

If I understand correctly, you need to have a pole on the unit circle, equal to \$0.1f_s\$. If this is true, then your only way to get that zero is by having a complex value, therefore:
$$z=\mathrm{e}^{j2\pi 0.1} \tag{2}$$
This means that your filter needs to be in quadrature, or you will need two signal paths. So, without knowing the purpose of your wanting to build such a filter, unless you can tolerate a complex valued zero, you can't have a zero on the unit circle anywhere else except at \$[-1,1]\$. That is, if I understood what you want, because it's verry fuzzy.

This is what the command zplane([1 exp(-0.2i*pi)], [1 -2 1]) shows.