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Below 1D argumentation also explains the 2D case. First consider the DTFT property for the pair $x[n] \longleftrightarrow X(e^{j\omega})$ $$e^{j\omega_0 n} \cdot x[n] \longleftrightarrow X(e^{j(\omega - \omega_0)})$$ Then recognise that $(-1)^n = e^{j \pi n}$ which yields: $$e^{j\pi n} \cdot x[n] \longleftrightarrow X(e^{j(\omega - \pi)})$$ The ...