Problem calculating the argument of a complex variable In Signals & Systems 2nd Ed. written by A. V. Oppenheim,
there is a result of Fourier transformation:
$ \begin{align} H (j \omega) = \frac{1 + (j \omega / \omega_{0})^2 - 2 j \zeta (\omega / \omega_0)}{1 + (j \omega / \omega_{0})^2 + 2 j \zeta (\omega / \omega_0)} \end{align} $
And from this, it states that the angle between the line from the origin to that point and the positive real axis in the complex plane (i.e. argument) is
$ \angle H (j \omega) = -2 \tan ^{-1} \left[ \frac{2 \zeta (\omega / \omega_{0})}{1 - (\omega / \omega_{0})^2} \right] $
where $\omega$ is angular frequency in physics, but it's not important. Just say $\omega$ is a real number. $j^2 = -1$.
I have a trouble deriving this.
Let's simplify these equations a little.
Let $x$ be $\frac{\omega}{\omega_{0}}$ in the previous equations.
Then $ \begin{align} H (x) = \frac{1 + (i x)^2 - 2 i \zeta x}{1 + (i x)^2 + 2 i \zeta x}, \tag{1} \end{align} $
and what I want to derive is $ \begin{align} \angle H (x) = -2 \tan ^{-1} \left[ \frac{2 \zeta x}{1 - x^2} \right]. \tag{2} \end{align} $
Here are what I calculated.
$ \mathcal{Re} \left\{ H (x) \right\} = \frac{(1-x^2)^2 - 4 \zeta^2 x^2}{(1 - x^2)^2 + 4 \zeta^2 x^2} $
$ \mathcal{Im} \left\{ H (x) \right\} = - \frac{4 \zeta x (1 - x^2)}{(1 - x^2)^2 + 4 \zeta^2 x^2} $
$ \begin{align}
\angle H (x) &= \tan ^{-1} \frac{\mathcal{Im} \left\{ H (x) \right\}}{\mathcal{Re} \left\{ H (x) \right\}} \\
&= \tan^{-1} - \frac{4 \zeta x (1 - x^2)}{(1-x^2)^2 - 4 \zeta^2 x^2} \\
&= - \tan^{-1} \frac{4 \zeta x (1 - x^2)}{(1-x^2)^2 - 4 \zeta^2 x^2}
\end{align} $
Using $ \tan^{-1} x = 2 \tan^{-1} \frac{x}{1 + \sqrt{1 + x^2} } $ (from Wikipedia),  
$ \begin{align}
\angle H (x) &= -2 \tan^{-1} \frac{x^2 - 1}{2 \zeta x}
\end{align} $
However, in the textbook,
$ \angle H (x) = -2 \tan^{-1} \frac{2 \zeta x}{1 - x^2} $
How come?
Is $ -2 \tan^{-1} \frac{x^2 - 1}{2 \zeta x} = -2 \tan^{-1} \frac{2 \zeta x}{1 - x^2} $ or did I miss something?

You have probably a mistake in your last calculation:
$$\alpha = -2 \tan^{-1} \left( \frac{y}{1 + \sqrt{1+y^2}}  \right) , \text{with } y = \frac{4 \zeta x(1-x^2)}{(1-x^2)^2 - 4 (\zeta x)^2}$$
I got
$$
\sqrt{1+y^2} = \frac{(1-x^2)^2 + 4(\zeta x)^2}{(1-x^2)^2-4(\zeta x)^2}
$$
and therefore
$$
1+\sqrt{1+y^2} = \frac{2(1-x)^2}{(1-x)^2-4(\zeta x)^2}.
$$
Hence
$$
\alpha = -2 \tan \left( \frac{ \frac{4 \zeta x(1-x^2)}{(1-x^2)^2 - 4 (\zeta x)^2} }{ \frac{2(1-x)^2}{(1-x)^2-4(\zeta x)^2} } \right) = \frac{2 \zeta x}{1-x^2}
$$
which is the angle you were looking for.
