How to find a transfer function when an \$\omega\$ is not paired with \$j\$?

Question

With the standard transfer function of:

$$H(s) = \frac{N(s)}{D(s)} = \frac{(s-z_1)(s-z_2)...(s-z_{m-1})(s-z_m)}{(s-p_1)(s-p_2)...(s-p_{n-1})(s-p_n)}$$

and defining: \$s = \sigma + j\omega\$

Can you put a transfer function into this form if you have an \$\omega\$ not multiplied by a \$j\$?

To give some context, you can run into this scenario when finding the impedance of a capacitor using a series R,L,C model.

^{simulate this circuit – Schematic created using CircuitLab}

$$\hat{Z} = R + j\omega L + \frac{1}{j\omega C}$$

$$\hat{Z} = \frac{j\omega RC - \omega^2LC + 1}{j\omega C}$$

If you treat \$\omega\$ as your input variable and \$\hat{Z}\$ as your output variable, then the complex impedance can be seen as a transfer function. For the equation shown, can you put this into the pole-zero format?

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Edit:

What I was not seeing here is that in \$s = \sigma + j\omega\$, \$\sigma==0\$ for sinusoidal inputs, as there is no real (decaying) part to the input (See Why do we use \$s=j\omega\$ in AC analysis instead of \$s=\sigma+j\omega\$?). This means that \$s^2 = -\omega^2\$. Thus my previous equation can be changed to look like:

$$\hat{Z} = L\frac{-\omega^2 + j\omega \frac{R}{L} + \frac{1}{LC}}{j\omega}$$ $$\hat{Z} = L\frac{s^2 + s\frac{R}{L} + \frac{1}{LC}}{s}$$

See Matt L.'s answer for the rest.

Answer 1

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.