given $Z_i ~ N(0,1)$ and $Z^2 = Z_1^2 + Z_2^2$ what is $Cov(Z^2,Z_1)$? Currently I am at the stage where:
$Cov(Z^2,Z_1) = E(Z^2*Z_1) - E(Z^2)E(Z_1)$
$=E(Z^2*Z_1)$, because expectations of normal, or combination of normal variables is zero.
After this I have no idea how to calculate the rest.
 A: Since you have not specified to the contrary, I will assume you intend that $Z_1 \perp Z_2$.  Replicating the steps you already have, and then expanding out the $Z^2$ term you get:
$$\begin{equation} \begin{aligned}
\mathbb{Cov}(Z^2, Z_1) = \mathbb{E}(Z^2 Z_1) 
&= \mathbb{E}(Z_1^3 + Z_1 Z_2^2) \\[6pt]
&= \mathbb{E}(Z_1^3) + \mathbb{E}(Z_1) \mathbb{E}(Z_2^2) \\[6pt]
&= \mathbb{E}(Z_1^3) \\[6pt]
&= 0. \\[6pt]
\end{aligned} \end{equation}$$
These steps only require that $Z_1 \perp Z_2$ and that $Z_1$ is symmetric around zero; you do not need to assume normality, or make any assumption about the variance.  So you can see that $Z_1$ and $Z^2$ are uncorrelated, so long as these underlying conditions hold.  An analogous result holds when we compare $Z_2$ and $Z^2$.  This result is telling you that, for independent variables that are symmetric around zero, the squared-norm of the vector of these variables is uncorrelated with the individual variables.  This is unsurprising, given the symmetric condition.
