If you read about the real part then you must be able to read about the imaginary part, as well, along with explanations. What you're asking can take a good part of a book.
Otherwise, to find out how the real- and imaginary parts affect a generic 2nd order biquad, write it this way:
$$\begin{align}
H(s)&=K\dfrac{a_2s^2+a_1s+a_0}{b_2s^2+b_1s+b_0} \tag{1} \\
&=K\dfrac{s^2-2\Re{}_z s+\Re{}_z^2+\Im{}_z^2}{s^2-2\Re{}_ps+\Re{}_p^2+\Im{}_p^2} \tag{2}
\end{align}$$
Where \$\Re\$ is considered negative for a Hurwitz polynomial. Now you can apply the inverse Laplace transform and you'll get:
$$\begin{align}
h(t)&=K\Big\{\delta(t)+\mathrm{e}^{\Re{}_pt}\big[A\sin(\Im{}_pt)+B\cos(\Im{}_pt)\big]\Big\} \tag{3} \\
A&=\dfrac{\Re{}_p^2-\Im{}_p^2+\Re{}_z^2+\Im{}_z^2-2\Re{}_p\Re{}_z}
{\Im{}_p} \\
B&=2(\Re{}_p-\Re{}_z)
\end{align}$$
If you look at the terms, closely, \$\Im{}_p\$ (the imaginary part of the pole) appears as the argument for the oscillating terms, \$\sin\$ and \$\cos\$. This implies an underdamped case, otherwise it would have been an overdamped case, \$\sinh\$ and \$\cosh\$. If none are present then it's critically damped.
Note that (1) differs from (2) in that \$K\$ is different for the two, which is reflected in the \$\delta(t)\$ term: for (1) it would have been \$K\frac{a_2}{b_2}\$, which means that the \$K\$ in (2) is, in fact: \$K\frac{\Re{}_z^2+\Im{}_z^2}{\Re{}_p^2+\Im{}_p^2}\$. But, since it's only a matter of scaling, I've considered it as one.
And for the "why" part, it helps if you think of what the real- and imaginary parts of a pole represent:
- \$s=\sigma\pm j\omega =\Re\pm j\Im\$ is the general representation of the pole;
- \$\Re\$ is on the X axis, the damping, \$\zeta\$ ,and the quality factor,\$Q\$, are dependent on it, and it appears as the argument for \$\mathrm{e}^{\Re t}\$, so it sets the decaying envelope;
- \$\Im\$ is on the imaginary axis, where the frequency lies (\$j\omega\$), so it sets the oscillation.