$az+b\overline{z}+c=0$ represents a line

Question

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?

Write out your expression in terms of the real and imaginary parts. — copper.hat, Sep 01 '13 at 18:06
http://ocw.mit.edu/courses/mathematics/18-112-functions-of-a-complex-variable-fall-2008/assignments/ps1.pdf — lab bhattacharjee, Sep 01 '13 at 18:10
Look, the expression above puts an affine constraint on the real and imaginary parts of $z$, so the only question is whether the set it describes is empty, a line or the whole space. — copper.hat, Sep 01 '13 at 18:11

Caleb Stanford · Accepted Answer · 2013-09-01T19:07:23.173

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.

$az+b\overline{z}+c=0$ represents a line

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