Nulling filter coefficients
You are looking at an FIR filter whose transfer function is 

$$
\begin{align}
H(z)&amp;=(1-e^{j\omega_0}z^{-1})(1-e^{-j\omega_0}z^{-1})\\[10pt]
&amp;=1-(e^{-j\omega_0}+e^{j\omega_0})z^{-1}+z^{-2}\\[10pt]
&amp;=1-2\cos(\omega_0)z^{-1}+z^{-2}
\end{align}$$

So as you can also see in the comment, there are two zeros at $e^{\pm j\omega_0}$ that null the frequencies $\pm\omega_0$.  

Obviously, here we have $\omega_0=0.44\pi$. The filter's response is 



The impulse response is $$h[n]=\delta[n]-2\cos(0.44\pi)\delta[n-1]+\delta[n-2]$$

Assume an arbitrary input $x[n]$. The output $y[n]$ is given by:

$$y[n]=x[n]*h[n]$$ Therefore $$y[n]=x[n]-2\cos(0.44\pi)x[n-1]+x[n-2]$$