Solving a complex quadratic-like equation Let $a,b,c,d$ are given non-zero complex numbers (i.e., constants). Then is it true that the equation 
$$a|z|^2+b z+c\bar{z}+d=0$$
which is equivalent to $$|z|^2+b'z+c'\bar{z}+d'=0$$
will always have a (at least one) solution for $z$? Or there is some necessary and sufficient condition that $a,b,c,d$ must satisfy to guarantee a solution in $z$? Notation: $\bar{z}$ is the complex conjugate of $z$.

The equation wil not always have solutions, as pointed out already.
The following gives a necessary condition for solutions to exist. Consider WLOG the case $\,a=1\,$, then taking the complex conjugates on both sides gives $\,|z|^2+\bar b \bar z+ \bar c z+\bar d=0\,$. Eliminating $\bar z\,$ between the latter and the original equation results in:
$$
(\bar b - c) |z|^2+(|b|^2-|c|^2) z + \bar b d - c \bar d = 0 \quad \iff \quad (|b|^2-|c|^2) z = (c - \bar b) |z|^2 + c \bar d - \bar b d
$$
Taking the squared modulus on both sides:
$$
(|b|^2-|c|^2)^2 |z|^2 = \big((c - \bar b) |z|^2 + c \bar d - \bar b d\big)\big((\bar c - b) |z|^2 + \bar c d - b \bar d\big)
$$
The latter is a quadratic in $\,|z|^2\,$, and a necessary condition for solutions to exist is that the quadratic must have at least one real positive root. This can be easily verified for particular values of $\,b, c, d\,$, but writing the condition in the general case in terms of arbitrary $\,b, c, d\,$ is not pretty.
Also, the condition is not necessarily sufficient: if a real positive root does exist, then its square root gives $\,|z|\,$, which can then be substituted back into the original equation to get $\,z\,$, but it still remains to be verified that the resulting $\,z\,$ does in fact satisfy the equation.
