How to find a transfer function when an \$\omega\$ is not paired with \$j\$?
As a function of \$s\$ you get for the impedance

$$Z(s)=R+sL+\frac{1}{sC}=\frac{s^2LC+RCs+1}{sC}=L\frac{s^2+\frac{R}{L}s+\frac{1}{LC}}{s}\tag{1}$$

Computing the zeros of the numerator of (1) gives

$$z_{1,2}=-\frac{R}{2L}\pm\sqrt{\frac{R^2}{4L^2}-\frac{1}{LC}}\tag{2}$$

With these zeros, the impedance (1) can be written as

$$Z(s)=L\frac{(s-z_1)(s-z_2)}{s}\tag{3}$$

From (3) it is clear that apart from the two zeros at \$s=z_1\$ and \$s=z_2\$ we have two poles at \$s=0\$ and \$s\rightarrow\infty\$, which makes sense because for DC the impedance becomes infinite due to the capacitor, and for \$\omega\rightarrow\infty\$ the impedance goes to infinity due to the inductor.