Put $X=[x_1,\cdots,x_N]^T$. We consider an equation in the form $X\circ X+A'+B'X+C'\overline{X}=0$ where $\circ$ is the Hadamard product, $A'$ is a vector and $B',C'$ are square matrices.

Step 1. Put, for every $k$, $x_k=1+y_k$. Then $Y$ satisfies an equation in the previous form. Then we may assume that $|x_k|\leq 1$.

Step 2. Separate real and imaginary parts with $X=U+iV$. We obtain a system (in the unknown REAL vectors $U,V$) in the following form $\{f(U,V)=U\circ U-V\circ V+A+BU+CV=0,\;\;g(U,V)=2U\circ V+D+EU+FV=0\}$.

Solving (exactly) this system is hopeless. For $N=2$, there are $2^4$ solutions in $\mathbb{C}$ (using the Grobner basis method, we must solve an irreducible polynomial of degree $2^4$). I think that, more generally, there are $2^{2N}$ solutions in $\mathbb{C}$. Obviously, we keep only the real solutions!

You search local solutions ; then you can use the Newton's method. Let $F=[f,g]^T$. The recurrence formula is well-known: $[U_{n+1},V_{n+1}]^T=[U_n,V_n]^T-(DF([U_n,V_n]^T))^{-1}([f(U_n,V_n),g(U_n,V_n)]^T$. Here $DF([U,V]^T)=\begin{pmatrix}2\tilde{U}+B&-2\tilde{V}+C\\2\tilde{V}+E&2\tilde{U}+F\end{pmatrix}$ where $\tilde{U}=diag(u_1,\cdots,u_N)$ and $\tilde{V}=diag(v_1,\cdots,v_N)$

and(quadratic) inequalities. – Dima Pasechnik May 31 '14 at 7:27