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

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 $$a\neq 0,\quad b=\bar{a},\quad c\in\Bbb R.$$

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).

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 $x-y+1=0$