Translation property of 2-D discrete Fourier transform

Question

I am studying the 2-D discrete Fourier transform related to image processing and I don't understand a step about the translation property.

In the book Digital Image Processing (Rafael C. Gonzalez, Richard E. Woods ) is written that the translation property is:

$$f(x,y)e^{j2\pi(\frac{u_0x}{M} + \frac{v_0y}{N})} \Leftrightarrow F(u-u_0, v-v_0)$$ and $$f(x-x_0,y-y_0) \Leftrightarrow F(u,v)e^{-j2\pi(\frac{x_0u}{M} + \frac{y_0v}{N})}$$

So if I want to translate the Fourier Transform in the center of the center of the frequency rectangle, I had to consider $u_0=M/2$ and $v_0=N/2$. So the first line becomes:

$$f(x,y)e^{j2\pi(\frac{M}{2}x\cdot\frac{1}{M} + \frac{N}{2}y\cdot\frac{1}{N})} \Leftrightarrow F(u-\frac{M}{2}, v-\frac{N}{2})$$

$$f(x,y)e^{j2\pi(\frac{x}{2} + \frac{y}{2})} \Leftrightarrow F(u-\frac{M}{2}, v-\frac{N}{2})$$

$$f(x,y)e^{j\pi(x+y)} \Leftrightarrow F(u-\frac{M}{2}, v-\frac{N}{2})$$

Then the book draws this conclusion: $$f(x,y)(-1)^{x+y} \Leftrightarrow F(u-\frac{M}{2}, v-\frac{N}{2})$$

Why? What are the steps it did? I know that $j$ is a complex number then $j^2=-1$.

Answer 1

Very informally, they use the property $e^{j\pi} = -1$ (which is correct), the property $e^{ab}={(e^{a})}^b$ (valid for certain values of $b$), and derive the final expression for $b=x+y$ when $x+y$ is an integer.

The pattern $(-1)^{x+y}$ looks like a checkerboard.