Solution:

For $N=3$, the transfer function of a normalized Butterworth filter is given by,

$\begin{aligned}\\ H(s) & =\frac{1}{(s+1)\left(s^2+s+1\right)} \\\\ & =\frac{A}{s+1}+\frac{B}{s+0.5+j 0.866}+\frac{C}{s+0.5-j 0.866}\\ \end{aligned}\\ $

$\begin{aligned} A & =\left.(s+1) \frac{1}{(s+1)\left(s^2+s+1\right)}\right|_{s=-1}=\frac{1}{(-1)^2-1+1}=1 \\\\ B & =\left.(s+0.5+j 0.866) \frac{1}{(s+1)(s+0.5+j 0.866)}\right|_{s=-0.5-j 0.866} \\\\ & =\frac{1}{(-0.5-j 0.866+1)(-j 0.866-j 0.866)} \\\\ & =\frac{1}{-j 1.732(0.5-j 0.866)}=\frac{1}{-j 0.866-1.5} \\\\ & =\frac{-1.5+j 0.866}{3}=-0.5+j 0.288 \\\\ C & =B^*=-0.5-j 0.288\\ \end{aligned}\\ $

Hence,

$\begin{aligned} H(s) & =\frac{1}{s+1}+\frac{-0.5+0.288 j}{s+0.5+j 0.866}+\frac{-0.5-0.288 j}{s+0.5-j 0.866} \\\\ & =\frac{1}{s-(-1)}+\frac{-0.5+0.288 j}{s-(-0.5-j 0.866)}+\frac{-0.5-0.288 j}{s-(-0.5+j 0.866)}\\ \end{aligned}$

In the impulse invariant technique,

if $H(s)=\sum_{k=1}^N \frac{c_k}{s-p_k}$, then $H(z)=\sum_{k=1}^N \frac{c_k}{1-e^{p_k T} z^{-1}}$

Therefore,

$\begin{aligned} H(z) & =\frac{1}{1-e^{-1} z^{-1}}+\frac{-0.5+j 0.288}{1-e^{-0.5} e^{-j 0.866} z^{-1}}+\frac{-0.5-j 0.288}{1-e^{-0.5} e^{j 0.866} z^{-1}} \\\\ & =\frac{1}{1-0.368 z^{-1}}+\frac{-1+0.66 z^{-1}}{1-0.786 z^{-1}+0.368 z^{-2}}\\ \end{aligned}$