Phase function of filter I have a filter with the transfer function 

$$H(z) = 1 - 2z^{-2} + z^{-4}.$$ The task is to find the phase function $\theta (\omega).$



My attempt is to start by expressing the frequency response 
\begin{align}
H(\omega) &amp;=H(z)\big\vert_{z=e^{j\omega}}\\
&amp;= 1 - 2e^{-2j\omega} + e^{-4j\omega}\\
&amp;=  e^{-2j\omega}( e^{2j\omega} - 2 + e^{-2j\omega})\\
&amp;=  e^{-2j\omega}(2\cos(2\omega)-2)
\end{align}

I think there is a relation of the sort $$H(\omega) = e^{j\theta (\omega)}|H(\omega)|$$

which in my case would give me $$\theta (\omega) = -2\omega.$$

The correct answer is $$\theta (\omega) = -2\omega + \pi.$$ How do I get the extra term?
HINT:

Answer these two basic questions, and you should be able to solve your problem:


What's the sign of the term $2\cos(2\omega)-2$? Can it be a magnitude?
What does a term of $\pm \pi$ in the phase do to the sign of an expression?