Deriving 2-D discrete Fourier transforms

Question

I have a problem in DFT. It was one of my past-year exam papers questions.

Question:

Let $F(u,v)$ be the 2-D Fourier transform of a 2-D continuous function $f(x,y)$. Derive in terms of $F(:,:)$ the 2-D Fourier transform of each of the following functions

1) $f(x,-2y)$

2) $f(x+2y,y)$

I know how to do 1-D Fourier transforms but not 2-D. I'm not sure how to start on it and need some guidance.

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For the second part, this was my approach. Please let me know if it's right or correct me if it's wrong.

Let $τ= x + 2y$ hence $x = τ-2y$ and $dx = dτ$ $$ \begin{align} \mathfrak{F}\{f(x+2y,y)\}&=∬ f(τ,y)e^{−j2π(u(τ-2y) +vy )} dx\ dv\\ \mathfrak{F}\{f(x+2y,y)\}&= ∬ f(τ,y)e^{−j2π(uτ+(-2u+v)y )} dx\ dτ \\ \mathfrak{F}\{f(x+2y,y)\}&= F(u,-2u+v) \end{align} $$

Answer 1

Here is the first one:

By definition,

$\mathfrak{F}\{ f(x,-2y)\} = \iint_{- \infty}^{\infty} f(x,-2y)e^{-j2\pi (ux+vy)}dxdy$

let $\tau = -2y$ and conversely $y=\frac{-\tau}{2}$

$\mathfrak{F}\{ f(x,-2y)\} = \iint_{- \infty}^{\infty} f(x,\tau)e^{-j2\pi (ux-\frac{v\tau}{2})}dxd(-\frac{\tau}{2})$

$\mathfrak{F}\{ f(x,-2y)\} = -\frac{1}{2}\iint_{- \infty}^{\infty} f(x,\tau)e^{-j2\pi (ux-\frac{v\tau}{2})}dxd\tau$

$\mathfrak{F}\{ f(x,-2y)\} = -\frac{1}{2} F(u,-\frac{v}{2})$

The other one will be harder, but I will leave it to you.