$a,b,c$ 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 $b\neq 0$, $t\in\mathbb{R}$

  • $\begingroup$ And what is the meaning of c then? $\endgroup$ – imranfat Apr 27 '13 at 22:03

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


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


and we can write $\frac{b}{a}=u^2$, where $|u|=1$. Now multiply by $\bar{u}$, to get


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).


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

  • $\begingroup$ Not really: if you multiply by an arbitrary nonzero number you don't get the relation $b=\bar{a}$. $\endgroup$ – egreg Apr 29 '13 at 17:50
  • $\begingroup$ @egreg: you are right. I have rectified this in an answer synthetizing both your approaches. $\endgroup$ – Georges Elencwajg Apr 29 '13 at 22:00

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 $x-y+1=0$

  • $\begingroup$ If you use an approach similar to leo's starting from the equation of a conic with center, you get a remarkably simple way to find the principal axes. $\endgroup$ – egreg Apr 29 '13 at 22:05
  • 1
    $\begingroup$ Try changing $x$ into $(z+\bar{z})/2$ and $y$ into $(z-\bar{z})/(2i)$ in the equation of a non-degenerate conic. You'll find something like $\bar{a}z^2+2bz\bar{z}+a\bar{z}^2+...$ with real $b$. If $a=0$ you have a circle; if $|a|^2=b^2$ you have a parabola. Otherwise it's easy to find the center and translate it to the origin; then you easily find a rotation around the origin that sends it to the form $Z^2+2BZ\bar{Z}+\bar{Z}^2+C=0$ which is nothing else than the normal form. $\endgroup$ – egreg Apr 29 '13 at 22:34

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: $$ \begin{align*} a_1x-a_2y+b_1x+b_2y&=-c_1 \\ a_1y+a_2x-b_1y+b_2x&=-c_2 \end{align*} $$


$$ \begin{align*} (a_1+b_1)x+(b_2-a_2)y&=-c_1 \\ (a_1-b_1)x+(b_2+a_2)y&=-c_2 \end{align*} $$


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