Parallel RLC tank transfer function
I believe the calculation for the first formula goes something like:

Using \$\omega_0^2 = (LC)^{-1}\$:

$$\begin{align}
Z(j\omega) &amp;= \frac{1}{\frac{1}{R} + \frac{1}{j\omega L} + j\omega C} \\
&amp;= \frac{j\omega RL}{j\omega L + R + j\omega RLC} \\
&amp;= \frac{j\left(\frac{\omega}{\omega_0}\right)\omega_0RL}{j\left(\frac{\omega}{\omega_0}\right)\omega_0L+ R\left(1-\left(\frac{\omega}{\omega_0}\right)^2\right)} \\
&amp;= \frac{j\omega_0RL}{j\omega_0L + R\left(\frac{\omega_0}{\omega} - \frac{\omega}{\omega_0}\right)}
\end{align}$$

We can linearize \$f(\omega) = \frac{\omega_0}{\omega} - \frac{\omega}{\omega_0}\$ around \$\omega_0\$ using a Taylor expansion:

$$f(\omega) \approx f(\omega_0) + \left.\frac{df}{d\omega}\right|_{\omega=\omega_0}(\omega - \omega_0) + ...$$

You can then find that

$$f(\omega) \approx -\frac{2}{\omega_0}(\omega - \omega_0)$$

Allowing us to continue with \$Z(j\omega)\$:

$$\begin{align}
Z(j\omega) &amp;= \frac{j\omega_0RL}{j\omega_0L + R\left(\frac{\omega_0}{\omega} - \frac{\omega}{\omega_0}\right)} \\
&amp;\approx \frac{j\omega_0RL}{j\omega_0L - \frac{2R}{\omega_0}(\omega - \omega_0)} \\
&amp;= \frac{R}{1 + j2\frac{R}{\omega_0L}\frac{\omega - \omega_0}{\omega_0}} \end{align}$$

Now we just use the definition for the Q-factor for parallel RLC circuits:

$$Q = R\sqrt{\frac{L}{C}} = \frac{R}{\omega_0L}$$

Note that this is not the same as the Q-factor of an individual inductor or capacitor, which is usually defined as \$Q_C = \frac{1}{\omega_0CR_S}\$ or \$Q_L = \frac{\omega_0L}{R_S}\$ where \$R_S\$ is the series resistance.

Plugging this into the previous equation will result in the equation you found.

The phase of \$Z(j\omega)\$ is found by using:

$$\begin{align}
\angle Z(j\omega) &amp;= \tan^{-1}\left(\frac{\mathcal{I}\{Z(j\omega)\}}{\mathcal{R}\{Z(j\omega)\}}\right) \\
&amp;= \frac{\pi}{2} - \tan^{-1}\left(\frac{\mathcal{R}\{Z(j\omega)\}}{\mathcal{I}\{Z(j\omega)\}}\right)
\end{align}$$

The second one is just using the trigonometric identity:

$$\tan\left(\frac{\pi}{2} - \alpha\right) = \cot(\alpha) = \frac{1}{\tan(\alpha)}$$

They probably used this version for their calculations.