Question: What are the conditions for which this way of solving hold?
I've a transfer function that looks like (in the Laplace domain):
$$\text{H}_\text{T}\left(\text{s}\right)=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}\tag1$$
Now, in the complex frequency domain it looks like (substitute $\text{s}=j\omega$ where $j^2=-1$):
$$\text{H}_\text{T}\left(j\omega\right)=\frac{\text{Y}\left(j\omega\right)}{\text{X}\left(j\omega\right)}\tag2$$
To solve the poles and zeros of $(1)$, we need to solve $\text{Y}\left(\text{s}\right)=0$ and $\text{X}\left(\text{s}\right)=0$. This can also be done by using the complex frequency domain (then we get the absolute value of the poles and zeros because $\left|\text{s}\right|=\left|j\omega\right|=\omega$):
$$\Re\left[\text{H}_\text{T}\left(j\omega\right)\right]=\Im\left[\text{H}_\text{T}\left(j\omega\right)\right]\tag3$$
But this does not hold for every transfer function. So I want to know and prove for which transfer fuctions this hold.
An example for where it does not hold:
$$\text{H}_{\text{T}}\left(\text{s}\right)=\frac{\text{R}_2+\text{s}\text{L}}{\text{R}_1+\text{R}_2+\text{s}\text{L}}\tag4$$
Where $\text{R}_1\space\wedge\space\text{R}_2\space\wedge\text{L}\in\mathbb{R}^+$.
As far as I could proceed:
$$\Re\left[\text{H}_\text{T}\left(j\omega\right)\right]=\Im\left[\text{H}_\text{T}\left(j\omega\right)\right]\space\Longrightarrow\space\Re\left[\text{Y}\left(j\omega\right)\cdot\overline{\text{X}\left(j\omega\right)}\right]=\Im\left[\text{Y}\left(j\omega\right)\cdot\overline{\text{X}\left(j\omega\right)}\right]\tag5$$