$X_{1},X_{2} \sim N(0,1)$ and are independent. Show $\frac{X_{1}}{X_{2}}$ and $\sqrt{X_{1}^{2}+X_{2}^{2}}$ are independent

Since $X_1,X_2$ are independent, the joint PDF is

$$\tag{1}f_{X_1,X_2}(x_1,x_2) = \frac{1}{\sqrt{2\pi}}e^{-x_1^2/2}\frac{1}{\sqrt{2\pi}}e^{-x_2^2/2} = \frac{1}{2\pi}e^{-(x_1^2 + x_2^2)/2}$$

For $(X_1,X_2) \in \mathbb{R}^2$ we have $(U,V) \in (-\infty, \infty)\times [0,\infty)$ under the mapping

$$U = \frac{X_1}{X_2}, \quad V = \sqrt{X_1^2 + X_2^2}$$

However, it is not one-to-one since points $(X_1,X_2)$ and $(-X_1,-X_2)$ have the same image. There is also a problem of how $U$ is defined when $X_2 = 0$, but this can be ignored by setting $U$ to be $0$ on this set of probability mass $0$.

The joint PDF of $U,V$ can be obtained as

$$\tag{2}f_{U,V}(u,v) = 2f_{X_1,X_2}(x_1(u,v),x_2(u,v)) |J(u,v)|,$$

where the factor of $2$ accounts for the two-to-one nature of the mapping and the Jacobian $J$ is obtained from the inverse mapping $x_1 = \frac{uv}{\sqrt{1+u^2}}, x_2 =\frac{v}{\sqrt{1 + u^2}} $ as

$$\tag{3}|J(u,v)| = \left|\frac{\partial(x_1,x_2)}{\partial(u,v)} \right|= \frac{\partial x_1}{\partial u }\frac{\partial x_2}{\partial v }-\frac{\partial x_1}{\partial v }\frac{\partial x_2}{\partial u } = \frac{v}{1 + u^2}$$

Substituting into (2) using (1) and (3) we obtain

$$f_{U,V}(u,v) = \frac1{\pi(1+u^2)}ve^{-v^2/2}$$

Since the joint PDF factors into a product of a function of $u$ and a function of $v$, the random variables $U$ and $V$ are independent.