Let $\mathbf{n}\sim\mathcal{CN}(\mathbf{0}_N,\nu^2\mathbf{I}_N)$ be a circularly symmetric complex Gaussian vector of size $N$. Note that the entries $n_i$ are i.i.d. Let $q=\sum_{i=1}^N n_i$ and $$\mathbf{u}=\frac{\overline{q}}{|q|}\mathbf{n}=\frac{\sum_{i=1}^N \overline{n_i}}{\left|\sum_{i=1}^N n_i\right|}\mathbf{n}\,.$$ I'm interested in computing $\mathbb{E}\left\{\sum_{i=1}^N u_i^2\right\}$, but it could also be useful to have any pointers for higher powers $r\geq2$ as well.
My attempt so far: developing the sum, \begin{align} \sum_{i=1}^N u_i^2 &=\frac{1}{\left|\sum_{j=1}^N n_j\right|^2} \sum_{i=1}^N \left(\sum_{j=1}^N \overline{n_j}\right)^2n_i^2=\frac{1}{\left|\sum_{j=1}^N n_j\right|^2} \sum_{i=1}^N \sum_{j=1}^N\sum_{k=1}^N \overline{n_j}\overline{n_k}n_i^2\\ &=\frac{1}{\left|\sum_{j=1}^N n_j\right|^2}\sum_{i=1}^N |n_i|^4 +\frac{1}{\left|\sum_{j=1}^N n_j\right|^2} \sum_{i=1}^N\sum_{\substack{j=1\\j\neq i}}^N\overline{n_j}^2n_i^2 \\ &\qquad\qquad+ \frac{1}{\left|\sum_{j=1}^N n_j\right|^2}\sum_{i=1}^N\sum_{\substack{j=1\\j\neq i}}^N\sum_{\substack{k=1\\k\neq i,j}}^N\overline{n_j}\overline{n_k}n_i^2 \end{align}
and here I am stuck when trying to compute expectation, in particular because of the ratio with the squared magnitude of the sum.
Additionally, the second and third terms look straightforward at first glance due to the independence of the $n_i$'s (which lead to think that we can factor out expectations of each factor in the numerator), but I am unsure if this is the case involving the ratio over squared magnitude. I tend to think that the factors of the summands are no longer independent, and I don't see a way to work around it.
The answer is $\nu^2$.
Let $n = x + \mathrm{i} y$. The PDF of $x_1$ is given by $$f(x_1) = \frac{1}{\sqrt{\pi \nu^2}}\mathrm{e}^{- \frac{x_1^2}{\nu^2}}, \quad x_1 \in \mathbb{R}.$$
Let $r := \sum_j x_j$ and $s := \sum_j y_j$. By symmetry, we have \begin{align*} \mathbb{E}\left[\sum_j u_j^2\right] &= N \mathbb{E}[u_1^2]\\[6pt] &= N \mathbb{E}\left[\frac{(r - \mathrm{i} s)^2(x_1 + \mathrm{i} y_1)^2}{r^2 + s^2}\right]\\[6pt] &= \mathrm{i}N \mathbb{E}\left[\frac{2x_1y_1(r^2 - s^2)}{r^2 + s^2}\right] + \mathrm{i}N \mathbb{E}\left[\frac{2rs(y_1^2 - x_1^2)}{r^2 + s^2}\right]\\[6pt] &\qquad + N\mathbb{E}\left[\frac{r^2x_1^2 - r^2y_1^2}{r^2 + s^2}\right] + N\mathbb{E}\left[\frac{s^2y_1^2 - s^2x_1^2}{r^2 + s^2}\right] + N\mathbb{E}\left[\frac{4rsx_1y_1}{r^2 + s^2}\right]\\[6pt] &= 2N\mathbb{E}\left[\frac{r^2x_1^2 - r^2y_1^2}{r^2 + s^2}\right] + \frac{1}{N}\mathbb{E}\left[\frac{4r^2s^2}{r^2 + s^2}\right] \tag{1} \end{align*} where we use (by symmetry) \begin{align*} \mathbb{E}\left[\frac{2x_1y_1(r^2 - s^2)}{r^2 + s^2}\right] &= 0, \\[6pt] \mathbb{E}\left[\frac{2rs(y_1^2 - x_1^2)}{r^2 + s^2}\right] &= 0, \\[6pt] \mathbb{E}\left[\frac{r^2x_1^2 - r^2y_1^2}{r^2 + s^2}\right] &= \mathbb{E}\left[\frac{s^2y_1^2 - s^2x_1^2}{r^2 + s^2}\right],\\[6pt] \mathbb{E}\left[\frac{4rsx_1y_1}{r^2 + s^2}\right] &= \mathbb{E}\left[\frac{4rs\cdot \frac{rs}{N^2}}{r^2 + s^2}\right]. \end{align*} Explanation:
i) Since $x_1, \cdots, x_N, y_1, \cdots, y_N$ are i.i.d., we have (swap $x$ and $y$) $$\mathbb{E}\left[\frac{2x_1y_1(r^2 - s^2)}{r^2 + s^2}\right] = \mathbb{E}\left[\frac{2y_1x_1(s^2 - r^2)}{s^2 + r^2}\right] = - \mathbb{E}\left[\frac{2x_1y_1(r^2 - s^2)}{r^2 + s^2}\right]$$ which results in $\mathbb{E}\left[\frac{2x_1y_1(r^2 - s^2)}{r^2 + s^2}\right] = 0$. The second one and the third are similar.
ii) We have $\mathbb{E}\left[\frac{4rsx_1y_1}{r^2 + s^2}\right] = \mathbb{E}\left[\frac{4rsx_iy_j}{r^2 + s^2}\right]$ for all $i, j$. Thus, $\mathbb{E}\left[\frac{4rsx_1y_1}{r^2 + s^2}\right] = \mathbb{E}\left[\frac{4rs\cdot \frac{1}{N^2}\sum_{i,j} x_i y_j}{r^2 + s^2}\right] = \mathbb{E}\left[\frac{4rs\cdot \frac{rs}{N^2}}{r^2 + s^2}\right]$.
Using the identity (for $p > 0$) $$\frac{1}{p} = \int_0^\infty \mathrm{e}^{-t p}\,\mathrm{d} t,$$ we have \begin{align*} &2N\mathbb{E}\left[\frac{r^2x_1^2 - r^2y_1^2}{r^2 + s^2}\right] + \frac{1}{N}\mathbb{E}\left[\frac{4r^2s^2}{r^2 + s^2}\right]\\[6pt] ={}& 2N\mathbb{E}\left[(r^2x_1^2 - r^2y_1^2)\int_0^\infty \mathrm{e}^{- t(r^2 + s^2)}\,\mathrm{d} t\right] + \frac{1}{N}\mathbb{E}\left[4r^2s^2 \int_0^\infty \mathrm{e}^{-t(r^2 + s^2)}\,\mathrm{d} t\right]\\[6pt] ={}& 2N\int_0^\infty \mathbb{E}\left[(r^2x_1^2 - r^2y_1^2) \mathrm{e}^{- t(r^2 + s^2)}\right]\,\mathrm{d} t + \frac{1}{N}\int_0^\infty \mathbb{E}\left[4r^2s^2 \mathrm{e}^{-t(r^2 + s^2)}\right]\,\mathrm{d} t. \tag{2} \end{align*}
Let $r_1 := x_2 + x_3 + \cdots + x_N$ and $s_1 := y_2 + y_3 + \cdots + y_N$. Then $r_1, s_1$ are i.i.d. with $r_1, s_1 \sim \mathcal{N}(0, (N-1)\nu^2/2)$. Also, $x_1, y_1, r_1, s_1$ are independent. We have \begin{align*} &\mathbb{E}\left[(r^2x_1^2 - r^2y_1^2) \mathrm{e}^{- t(r^2 + s^2)}\right]\\[6pt] ={}& \int_{\mathbb{R}^4} \Big((x_1 + r_1)^2x_1^2 - (x_1 + r_1)^2y_1^2\Big) \mathrm{e}^{- t(x_1 + r_1)^2 - t (y_1 + s_1)^2}\frac{1}{\sqrt{\pi \nu^2}}\mathrm{e}^{- \frac{x_1^2}{\nu^2}} \frac{1}{\sqrt{\pi \nu^2}}\mathrm{e}^{- \frac{y_1^2}{\nu^2}}\\[6pt] &\qquad \cdot \frac{1}{\sqrt{\pi (N - 1) \nu^2}}\mathrm{e}^{- \frac{r_1^2}{(N - 1)\nu^2}}\frac{1}{\sqrt{\pi (N - 1) \nu^2}}\mathrm{e}^{- \frac{s_1^2}{(N - 1)\nu^2}} \,\mathrm{d}x_1 \,\mathrm{d} y_1\, \mathrm{d} r_1\, \mathrm{d}s_1\\[6pt] ={}& \frac{v^4}{2(N t\nu^2 + 1)^3}.\tag{3} \end{align*} Thus, we have \begin{align*} &2N\int_0^\infty \mathbb{E}\left[(r^2x_1^2 - r^2y_1^2) \mathrm{e}^{- t(r^2 + s^2)}\right]\,\mathrm{d} t\\[6pt] ={}& 2N\int_0^\infty \frac{v^4}{2(N t\nu^2 + 1)^3}\,\mathrm{d} t \\[6pt] ={}& \frac12 \nu^2. \tag{4} \end{align*}
Note that $r, s$ are i.i.d. with $r, s \sim \mathcal{C}(0, N\nu^2/2)$. We have \begin{align*} &\mathbb{E}\left[4r^2s^2 \mathrm{e}^{-t(r^2 + s^2)}\right]\\[6pt] ={}& \int_{\mathbb{R}^2} 4r^2s^2 \mathrm{e}^{-t(r^2 + s^2)}\frac{1}{\sqrt{\pi N \nu^2}}\mathrm{e}^{- \frac{r^2}{N\nu^2}}\frac{1}{\sqrt{\pi N \nu^2}}\mathrm{e}^{- \frac{s^2}{N\nu^2}}\,\mathrm{d}r \, \mathrm{d} s\\[6pt] ={}& \frac{N^2 v^4}{(Nt \nu^2 + 1)^3}. \tag{5} \end{align*} Thus, we have $$\frac{1}{N}\int_0^\infty \mathbb{E}\left[4r^2s^2 \mathrm{e}^{-t(r^2 + s^2)}\right]\,\mathrm{d} t = \frac{1}{N}\int_0^\infty \frac{N^2 v^4}{(Nt \nu^2 + 1)^3}\,\mathrm{d} t = \frac12 \nu^2. \tag{6}$$
From (1), (2), (4) and (6), we have $$\mathbb{E}\left[\sum_j u_j^2\right] = \nu^2.$$
We are done.