Solution:

$ H_a(s)=\frac{1 / 2}{s+0.1-j 3}+\frac{1 / 2}{s+0.1+j 3}\\ $

Then the transfer function of the digital filter

$ \begin{aligned} & H(z)=\frac{1 / 2}{1-e^{-0.1 T} e^{j 3 T} z^{-1}}+\frac{1 / 2}{1-e^{-0.1 T} e^{j 3 T} z^{-1}} \\\\ & H(z)=\frac{1-\left(e^{-0.1 T} \cos 3 T\right) z^{-1}}{1-\left(2 e^{-0.1 T} \cos 3 T\right) z^{-1}+e^{-0.2 T} z^{-2}}\\ \end{aligned}\\ $