How to find the impulse response of this system with complex roots?
It turns out that those terms with complex conjugate scalers and exponentials can be combined.  Rather than do it to your example, I would rather do it in the Z-domain.  Let

$$\begin{align}
H(z) &amp;= \frac{A_1}{1-p_1\,z^{-1}} + \frac{A_2}{1-p_2\,z^{-1}} \\
\\
&amp;= \frac{A_1(1-p_2\,z^{-1})+A_2(1-p_1\,z^{-1})}{(1-p_1\,z^{-1})(1-p_2\,z^{-1})}  \\
\\
&amp;= \frac{(A_1-A_1 p_2\,z^{-1})+(A_2-A_2 p_1\,z^{-1})}{1-(p_1+p_2)\,z^{-1} + p_1 p_2\, z^{-2}}  \\
\\
&amp;= \frac{(A_1+A_2) - (A_1 p_2 + A_2 p_1)z^{-1}}{1-(p_1+p_2)\,z^{-1} + p_1 p_2\, z^{-2}} \\
\\
\end{align}$$

Now it turns out that there are two ways for the impulse response, $h[n] = \mathcal{Z}^{-1} \{H(z)\}$, to be total real (with no imaginary component).  One way is for both poles $p_1,p_2$ to be real and that will also require the two coefficients $A_1,A_2$ to also be real.

$$\begin{align}
h[n] &amp;= \mathcal{Z}^{-1}\{H(z)\} \\
\\
 &amp;= \mathcal{Z}^{-1}\left\{ \frac{A_1}{1-p_1\,z^{-1}} \right\} + \mathcal{Z}^{-1}\left\{ \frac{A_2}{1-p_2\,z^{-1}} \right\} \\
\\
&amp;= A_1 p_1^n u[n] + A_2 p_2^n u[n] \\
\end{align}$$

The other way for $h[n]$ to be real is for both poles $p_1,p_2$ to be complex-conjugate of each other and that will also require the two coefficients $A_1,A_2$ to also be complex-conjugate.

$$\begin{align}
h[n] &amp;= \mathcal{Z}^{-1}\{H(z)\} \\
\\
 &amp;= \mathcal{Z}^{-1}\left\{ \frac{(A_1+A_2) - (A_1 p_2 + A_2 p_1)z^{-1}}{1-(p_1+p_2)\,z^{-1} + p_1 p_2\, z^{-2}} \right\}  \\
\end{align}$$

(i need to return to this later.)