When does $az + b\bar{z} + c = 0$ represent a line? $a,b,c$ and $z$ are all complex numbers. My idea was to show that it passes through the point $\infty$ in the extended complex plane, but I'm not quite sure how to execute that.
Update:
It says in the text that a straight line can be represented by a parametric equation $z = a+bt$, where $a$ and $b$ are complex numbers, and $b\neq 0$, $t\in\mathbb{R}$
 A: For the record let me fuse the two remarkable answers by leo and egreg here (I am making this answer community wiki and have upvoted their answers, of course).
 I challenge users to find it stated in the literature !
Theorem (egreg,leo)
Given complex numbers $a,b,c$ the following are equivalent:    
1) The equation $az+b\bar z+c=0$ represents a real affine line in $\mathbb C$.
2) There exist $\alpha, \beta\in \mathbb C^*,\; r\in \mathbb R$ such that $az+b\bar z+c=\alpha(\beta z+\bar \beta \bar z+r)$  .
3) $|a|=|b|\neq 0$ and $b\bar{c}=\bar{a}c$
For example, the equation $z-i\bar z+1-i=0$ represents a real affine line: which one ?
Answer: the line $x-y+1=0$
A: All points satisfying $az+b\bar{z}+c=0$ satisfy also $\bar{b}z+\bar{a}\bar{z}+\bar{c}=0$. Multiply the first by $\bar{a}$ and the second by $b$, getting
\begin{cases}
a\bar{a}z+\bar{a}b\bar{z}+\bar{a}c=0\\
b\bar{b}z+\bar{a}b\bar{z}+b\bar{c}=0
\end{cases}
Subtract to get
$$(a\bar{a}-b\bar{b})z=b\bar{c}-\bar{a}c$$
So there are infinite points satisfying the original equation only if $|a|=|b|$ and $b\bar{c}=\bar{a}c$.
Are these conditions also sufficient? Yes.
Suppose we have $az+b\bar{z}+c=0$ with $|a|=|b|\ne0$ and $b\bar{c}=\bar{a}c$. This represents the same set of points represented by
$$z+\frac{b}{a}\bar{z}+\frac{c}{a}=0$$
and we can write $\frac{b}{a}=u^2$, where $|u|=1$. Now multiply by $\bar{u}$, to get
$$\bar{u}z+u\bar{z}+\frac{c\bar{u}}{a}=0$$
I claim that $\frac{c\bar{u}}{a}$ is real:
$$
\frac{\overline{c\bar{u}}}{\bar{a}}=
\frac{\bar{c}u}{\bar{a}}=
\frac{b\bar{c}u}{b\bar{a}}=
\frac{\bar{a}cu}{b\bar{a}}=
\frac{cu}{b}=
\frac{c}{b}u^2\bar{u}=
\frac{c}{b}\frac{b}{a}\bar{u}=
\frac{c\bar{u}}{a}
$$
Therefore we have reduced the equation to
$$
\bar{u}z+u\bar{z}+C=0
$$
with a real $C$, and this is readily shown to be the equation of a line (see, for instance, leo's answer).
A: From the parametric complex equation of a line $L$
$$z=a+bt,\quad b\neq 0,\ t\in\Bbb R$$
it is clear that a point $z\in \Bbb C$, $$z=x+iy,\quad x,y\in \Bbb R$$ belongs to that line if and only if the point $(x,y)$ belongs to the line
$$(\Re a, \Im a) +(\Re b, \Im b)t,\quad t\in \Bbb R$$
in $\Bbb R^2$.
But then, there are real numbers $A,\ B,\ C$, $A$ and $B$ not both $0$, such that the equation of this line in $\Bbb R^2$ is
$$2Ax+2By=C.$$
So, a complex number $z$ belongs to $L$ if and only if 
$$2A\frac{z+\bar{z}}{2}+2B\frac{z-\bar{z}}{2i}=C.$$
After the computations, you get that this last equation is equivalent to
$$(A-Bi)z+(A+Bi)\bar{z}-C=0,$$
with $(A-iB)\neq 0$. Therefore
$$az+b\bar{z}+c=0$$
represents a line if
$$a\neq 0,\quad b=\bar{a},\quad c\in\Bbb R.$$
A: Hint: Using $a=a_1+ia_2$, $b,c$ defined the same and $z=x+iy$
$az+b\bar{z}+c=0$ can be rewritten as:
$$
\begin{align*}
a_1x-a_2y+b_1x+b_2y&=-c_1 \\
a_1y+a_2x-b_1y+b_2x&=-c_2
\end{align*}
$$ 
or
$$
\begin{align*}
(a_1+b_1)x+(b_2-a_2)y&=-c_1 \\
(a_1-b_1)x+(b_2+a_2)y&=-c_2
\end{align*}
$$
A: Suppose $az + b\bar{z} + c = 0$. So $\bar{a}\bar{z} + \bar{b}z + \bar{c} = \bar{b}z +\bar{a}\bar{z} + \bar{c} = 0$. So $az + b\bar{z} + c $ and $\bar{b}z +\bar{a}\bar{z} + \bar{c}$, are the same line iff there are infinitely any solutions for $z$, perpendicular lines iff $z$ is a specific point, and parallel lines if there are no solutions.  Multiplying both sides of the first equation by  $\bar{a}$ we gent $\bar{a}az + \bar{a}b\bar{z} + \bar{a}c = 0$. 
Multiplying both sides of the second equation by $b$ we get $b\bar{b}z + b\bar{a}\bar{z} + b\bar{c} = 0$. So $$|a|^2z + \bar{a}b\bar{z} + \bar{a}c = 0$$
$$|b|^2 z + \bar{a}b\bar{z} + b\bar{c} = 0$$
Subtracting one equation from the other gives us $(|a|^2 - |b|^2)z + \bar{a}c - b\bar{c} = 0$. Hence $(|a|^2 - |b|^2)z = b\bar{c} - \bar{a}c$. So if $|a|^2 - |b|^2 \neq 0$, then we can divide both sides by $|a|^2 - |b|^2$ and get $z = \frac{b\bar{c} - \bar{a}c}{|a|^2 - |b|^2}$. But then $z$ is a point of intersection, and hence our equation represents intersecting lines. Therefore, $az b\bar{z} + c$, is a line only if  $|a|^2 = |b|^2$, and hence $|a| = |b|$. 
But then $(|a|^2 -|b|^2)z + \bar{a}c - b\bar{c} = 0$, hence $\bar{a}c - b\bar{c} = 0$. Since there is no way $\bar{a}c - b\bar{c} = 0$, get us that $z$ is equal to a point, we know as long as $|a| = |b|$, we do not have intersecting lines. Since  there are multiple solutions to there are multiple solutions to $\bar{a}c = b\bar{c}$, we have that or two equations can't represent two different  parallel lines if $\bar{a}c = b\bar{c}$. 
So $az + b\bar{z} + c$ is a line iff $|a| = |b|$ and $\bar{a}c = b\bar{c}$.
