$az+b\overline{z}+c=0$ represents a line 
When does $az+b\overline{z}+c=0$ represent a line?

All of $a,b,c$ are complex numbers.
I know that a line in the complex plane is usually represented by $z=a+bt$, where the parameter $t$ runs through all real values. 
So suppose we're given the equation $az+b\overline{z}+c=0$. If $z$ represents a line $d+et$, we must have $a(d+et)+b(\overline{d}+\overline{e}t)+c=0$. Since the equation holds for all real values $t$, it must be that $ae+b\overline{e}=0$, and as a result $ad+b\overline{d}+c=0$. How can we get to a condition involving only $a,b,c$ from here?
 A: Your approach is a good idea, but let's instead just write $z = x + iy$.  Then
$az + b\overline{z} + c = 0$ becomes
\begin{align*}
a(x + iy) + b(x - iy) + c &= 0 \\
\iff (a + b)x + (ai - bi)y + c &= 0
\end{align*}
At this point we have what looks a lot like a basic line in $\mathbb{R}^2$ of the form $Ax + By + C = 0$, with one problem: $A,B,$ and $C$ are complex.  We want this line to be an actual line in $\mathbb{R}^2$, which means we want to be able to reduce it to $A'x + B'y + C' = 0$, with $A', B', C'$ real.  In other words, we need to be able to multiply $Ax + By + C$ by a constant nonzero (complex) number to make all the coefficients real.
Excluding the case $C = 0$ ($c = 0$), such a number exists if and only if $A$ and $B$ are both real multiples of $C$.  So we have the necessary and sufficient conditions
\begin{align*}
\frac{A}{C} &= \frac{a + b}{c} \in \mathbb{R} \\
\frac{B}{C} &= \frac{ai - bi}{c} \in \mathbb{R}
\end{align*}
That is, $az + b\overline{z} + c, \; c \ne 0$ is a line if and only if $\frac{a + b}{c},\frac{ai - bi}{c} \in \mathbb{R}$.
You should be able to deal with the case $c = 0$ in a similar fashion.
