$a \bar{z}z+ b\bar{z} + \bar{b}z+c=0$ - What curve it represents in the complex plane? Let $a,c \in \mathbb{R}$ and $b \in \mathbb{C}$. We consider the equation $$a \bar{z}z+ b\bar{z} + \bar{b}z+c=0.$$ What curve it represents in the complex plane?
I think it a circle, but I am not able to conclude. Any help?
 A: Writing $z=x+iy$, $b=\alpha+i\beta$ and noticing that
$$
b\bar z+\bar b z=b\bar z+\overline{b \bar z}=2\Re(b\bar z)
$$ your equation turns as follows
\begin{align*}
&az\bar z+b\bar z+\bar b z+c=0\;\;\\
\Longleftrightarrow\;\;&a(x^2+y^2)+2\Re[(\alpha+i\beta)(x-iy)]+c=0\\
\Longleftrightarrow\;\;&a(x^2+y^2)+2\alpha x+2\beta y+c=0
\end{align*}
which is the equation of a circle if $a\neq0$, of a line otherwise.
A: This is a case where writing $z=x+iy$ for $x,y$ real helps.
Let $b=b_x+i b_y$.
If $a=0$, then the equation is $xb_x + y b_y + c = 0$ which is easily seen to be a line.
If $a \neq 0$, we can take $a=1$ without loss of generality. Then the
equation is $x^2+y^2 +xb_x + y b_y + c = 0$. To find a simpler expression for
the quadratic $(x,y) \mapsto x^2+y^2 +xb_x + y b_y + c$, we find the minimum
at $(x_0,y_0) = - {1 \over 2} (b_x,b_y)$ and re write the equation in terms
of $(x_0,y_0)$ to get
$x^2+y^2 +xb_x + y b_y + c = (x-x_0)^2 + (y-y_0)^2 + c - (x_0^2+y_0^2)$.
Hence $x^2+y^2 +xb_x + y b_y + c = 0$ describes a circle iff
$c \le {1 \over 4} |b|^2$ and describes $\emptyset$ otherwise :-).
A: It represents a conic where a=b and h=0, ie, a circle. It may also represent a pair of straight lines.
A: If $a\ne0$, write $B=-b/a$ and $C=c/a$, so the equation is
$$
z\bar{z}-B\bar{z}-\bar{B}z+B\bar{B}=B\bar{B}-C
$$
that can be written
$$
|z-B|^2=B\bar{B}-C
$$
which is a circle, with center $B$ and radius $B\bar{B}-C$, if $B\bar{B}-C>0$, the point $B$ if $B\bar{B}-C=0$, the empty set if $B\bar{B}-C<0$.
If $a=0$ and $b=0$, the equation is either the plane (if $c=0$) or the empty set (if $c\ne0$), so we can assume $a=0$ and $b\ne0$. Write $z=x+iy$, so we get
$$
b(x-iy)+\bar{b}(x+iy)+c=0
$$
that becomes
$$
(b+\bar{b})x-i(b-\bar{b})y+c=0
$$
Since $b+\bar{b}=A$ is real and $b-\bar{b}=iB$ is purely imaginary, we get
$$
Ax+By+c=0
$$
and at least one among $A$ and $B$ is nonzero. This is the equation of a straight line.
