Fourier transform of free resolvent kernel in three dimensions

The free resolvent in $\mathbb{R}^3$ has this rapresentation: $$(R_0(z)f)(x)=\int_{\mathbb{R}^3}\frac{e^{i\sqrt{z}|x-y|}}{4\pi|x-y|}f(y)dy$$ with $\Im\sqrt{z}>0$. So its integral kernel is $$K(x,y)=\frac{e^{i\sqrt{z}|x-y|}}{4\pi|x-y|}$$ What is its Fourier transform with respect to $x$? According to my calculation it should be $$\frac{e^{-iy\cdot\xi}}{|\xi|^2-z}$$ Is it right?

-