Difference equation with complex zero
What you actually want to do is the following: you need a zero at $\omega=\pi/4$, but you also need to place a zero at $\omega=-\pi/4$ in order to get real-valued filter coefficients. So your transfer function becomes

$$\begin{align}H(z)&amp;=\frac{(z-e^{j\pi/4})(z-e^{-j\pi /4})}{z^2}\\&amp;=\frac{z^2-2z\cos(\pi/4)+1}{z^2}\\&amp;=1-\sqrt{2}z^{-1}+z^{-2}\tag{1}\end{align}$$

The corresponding difference equation is

$$y[n]=x[n]-\sqrt{2}x[n-1]+x[n-2]\tag{2}$$