$z_1 = \sqrt{-2\ln(x_1)}\cos(2\pi x_2)$ and $z_2 = \sqrt{-2\ln(x_1)}\sin(2\pi x_2)$ are standard normal random variables Context: http://mathworld.wolfram.com/Box-MullerTransformation.html
$x_1$ and $x_2$ are uniformly and independently distributed between $0$ and $1$.
The article says that it can be verified that $z_1$ and $z_2$ are independent standard normal random variables, and that this can be verified by solving for $x_1$ and $x_2$.
I can follow how they've solved for $x_1$ and $x_2$, but how does that show that $z_1$ and $z_2$ are standard normal variables?
 A: They explain this further down in the article. They take the Jacobian and get $$ \frac{\partial(x_1,x_2)}{\partial(z_1,z_2)} = \frac{1}{\sqrt{2\pi}}e^{-z_1^2/2}\frac{1}{\sqrt{2\pi}}e^{-z_2^2/2}=\phi(z_1)\phi(z_2) $$ where $\phi$ is the standard normal PDF. This means that, since the density of the uniforms $f_{X_1,X_2}(x_1,x_2)$ is just $1,$ the density of $(z_1,z_2)$ is $$f_{Z_1,Z_2}(z_1,z_2) = f_{X_1,X_2}(x_1(z_1,z_2),x_2(z_1,z_2))\left|\frac{\partial(x_1,x_2)}{\partial(z_1,z_2)}\right|= \phi(z_1)\phi(z_2)$$
i.e. they are independent standard normals.
It's instructive to also view the Box-Mueller from an intuitive angle. In polar coordinates two independent gaussians has density $$e^{-(x^2+y^2)/2}dxdy \sim re^{-r^2/2}drd\theta.$$ So $\theta$ is uniform on $[0,2\pi]$ and $r$ has density proportional to $re^{-r^2/2}dr \propto e^{-r^2/2}dr^2$ which by a change of variables means $r^2$ is exponential with mean $2.$ Now note that we can write the transformation $$(z_1,z_2) = (r\cos(\theta),r\sin(\theta))$$ where $$r^2 = -2\ln(u_1) \\ \theta = 2\pi u_2$$ and $u_1,u_2$ are independent uniforms on $[0,1]$. Note that $-2\ln(u_1)$ is an exponential with mean $2$ (this is a standard inverse CDF transformation) and $2\pi u_2$ is uniform on $[0,2\pi].$ So the two seed variables $u_1,u_2$ are related to the polar coordinates in a simple way. 
