Distribution of ratio between complex Gaussian and Chi-square R.V.s What would be the distribution of the following ratio?
$z = \frac{x_{1}}{|x_{1}|^2 + |x_{2}|^2 + ... + |x_{M}|^2}$
where $x_{i} \sim \mathcal{CN}(0,a), \forall i$ and $a > 1$. As can be seen, the denominator follows a Chi-square distribution with $2M$ degrees of freedom as $x_{i}$ are i.i.d. R.V.s. 
Remark 1: I've run some simulations in Matlab and it seems that the resulting distribution is Gaussian, or at least, it has a bell-shaped histogram.
 A: In the real Normal case, there is a partly closed-form resolution for the density, although there may be better expressions.
First, inverting $y=x_1/x_1^2+\xi$ in $x_1$ leads to two solutions
$$x_1^i(y) = \dfrac{1 \pm \sqrt{1-4y^2\xi}}{2y}\qquad i=1,2$$when $|y|<1/2\sqrt{\xi}$ and none when $|y|>1/2\sqrt{\xi}$.
Second, the distribution of $y$ is symmetric around zero and hence it is sufficient to consider $\mathbb{P}(Y\ge y_0)$ for $0<y_0<1/2\sqrt{\xi}$, for which we have
\begin{align*}
\mathbb{P}(Y\ge y_0) &= \mathbb{E}_\xi\left[\mathbb{P}\{
X_1\in (x_1^i(y_0),x^i_2(y_0))|\xi\}\right]\\
&= \mathbb{E}_\xi\left[\Phi(x^2_1(y_0))-\Phi(x^1_1(y_0))\right]\\
\end{align*}
where $\Phi$ ($\varphi$, resp.) denotes the Normal cdf (pdf, resp.). Therefore the density of $Y$ is given by
\begin{align*}
g(y_0) &= -\mathbb{E}_\xi\left[ \varphi(x^2_1(y_0))\frac{\text{d}x^2_1(y_0)}{\text{d}y_0}-\varphi(x^1_1(y_0))\frac{\text{d}x^1_1(y_0)}{\text{d}y_0} \right]
\end{align*}
where
$$\frac{\text{d}x^i_1(y_0)}{\text{d}y_0}=\frac{-1}{4y_0}^2
\left\{ 4y_0\left[1\pm \sqrt{1-4y_0^2\xi}\right]
\pm 8\xi/\sqrt{1-4y_0^2\xi}\right\}$$
