Conjugate symmetry of real-coefficient filters in Oppenheime's Discrete Time Signal Processing
The Fourier transform of a real function $x[n]$ is conjugate symmetric. That is,
$$X^*(\omega)=X(-\omega)$$
This is easy to observe. From the definition of FT: $$X(\omega)=\sum_{n=-\infty}^{+\infty}x[n]e^{-j\omega n}$$
Conjugating both sides:

$$\begin{align}
X^*(\omega)=&amp;\sum_{n=-\infty}^{+\infty}x^*[n]\left(e^{-j\omega n}\right)^*\\
&amp;=\sum_{n=-\infty}^{+\infty}x[n]e^{j\omega n}\\
&amp;=X(-\omega)
\end{align}$$

A direct consequence of this property (that you can easily verify) is that


  1- The real part and the magnitude of the Fourier transform of a real valued function are even functions of $\omega$. That is, $\Re\{X(\omega)\}=\Re\{X(-\omega)\}$ and $|X(\omega)|=|X(-\omega)|$.
  
  2- The imaginary part and the phase of the Fourier transform of a real valued function are odd functions of $\omega$. That is, $\Im\{X(\omega)\}=-\Im\{X(-\omega)\}$ and $\angle X(\omega)=-\angle X(-\omega)$.


Based on the above explanation, for a real-coefficient filter $h$, the magnitude of the frequency response is an even function.

$$\begin{align}
|H(e^{j(\omega-\pi)})|&amp;=|H(e^{j(-\omega-\pi)})|\\
&amp;=|H(e^{j(-\omega-\pi+2\pi)})|=|H(e^{j(\pi-\omega)})|
\end{align}$$
and the second line is because DTFT is $2\pi$-periodic. Hence we can add $2\pi$ and the result does not change.