Why does circular symmetry of complex envelopes mean that $\mathbb{E} \Big[\mathbf{x}[i] \mathbf{x}^T[i] \Big] = \mathbf{0}$?
A complex random variable with dimension being 2 $Z=X+jY$ can be defined as a Random vector with components $X$ and $Y$ which take values along 2 different dimensions. The Probability Density Function of $Z$ is defined using the PDFs of $X$ and $Y$. The Expectation is defined as $\mathrm{E}(Z)=\mathrm{E}(X)+j\mathrm{E}(Y)$ and the variance is defined the usual way as $Z$ being a complex function of $X$ and $Y$ and the Variance is $\mathrm{VAR}(Z)=\mathrm{E}(|Z^2|)-|\mathrm{E}(Z)|^2$. Intuitively speaking, $Z$ is a complex function of two random variables which just assumes value $X+jY, \forall x,\forall y$. The PDF is written as follows: Write the 2-dimensional Random variable as $$\tilde{Z}=\left[\begin{array}{ccc}X \\Y\end{array}\right]$$where $\tilde{Z}$ is a Real random vector. The PDF of $Z$ is the PDF of $\tilde{Z}$.
Now with Complex Random vectors, $\mathbf{Z}=\mathbf{X}+j\mathbf{Y}$ is a vector of Complex Random variables which has a lot of interesting properties in a sense that it has more to offer than a Vector of Real Random Variables. Particularly of importance are the Covariance, Pseudocovariance(this is something new) and the Expectation. In our situation, we have a special Complex Random vector called the Symmetric  Random Vector $\mathbf{Z}$.
Define the $2n$-dimensional Complex Random Vector as $$\tilde{\mathbf{Z}}=\left[\begin{array}{ccc}\mathbf{X} \\\mathbf{Y}\end{array}\right]$$ where $\mathbf{X}$ and $\mathbf{Y}$ are $n$-dimensional real random vectors. The PDF definition follows the same from Complex Random Variable. You can look up what the Pseudocovariance and Variance mean for a Complex Random Vector. I'm referring to Digital Communications by Proakis here in my definitions and notations. 

Now as in the comments, if $\mathbf{Z}$ is a Symmetric i.e the PDF doesn't change on rotation, we can say that $\mathrm{E}(\mathbf{ZZ^\mathrm{T}})=0$. This is a consequence of the following. Since $\mathbf{Z}$ and $\mathbf{Z}e^{j\theta}$ have the same PDF, $$\mathrm{E}(\mathbf{Z})=\mathrm{E}(e^{j\theta}\mathbf{Z})=e^{j\theta}\mathrm{E}(\mathbf{Z}), \forall \theta$$ If we had $\theta=\pi$ (the condition has to be satisfied for all $\pi$), we can see that $$\mathrm{E}(\mathbf{Z})=\mathrm{E}(-\mathbf{Z})=-\mathrm{E}(\mathbf{Z})$$ which clearly says that $\mathrm{E}(\mathbf{Z})=0$. Going further, we can prove that $\mathrm{E}(\mathbf{Z}\mathbf{Z}^\mathrm{T})=0$ the same way assuming a special case of $\theta=\frac{\pi}{2}$. And hence Circularly Symmetric Complex Random Vectors are Zero-mean and Proper. The last equation left unproven here is the definition of being proper.