Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
And what is the meaning of c then? – imranfat Apr 27 '13 at 22:03
up vote 6 down vote accepted

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).

share|cite|improve this 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 $$a\neq 0,\quad b=\bar{a},\quad c\in\Bbb R.$$

share|cite|improve this answer
Not really: if you multiply by an arbitrary nonzero number you don't get the relation $b=\bar{a}$. – egreg Apr 29 '13 at 17:50
@egreg: you are right. I have rectified this in an answer synthetizing both your approaches. – 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 $x-y+1=0$

share|cite|improve this answer
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. – egreg Apr 29 '13 at 22:05
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. – 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: $$ \begin{align*} a_1x-a_2y+b_1x+b_2y&=-c_1 \\ a_1y+a_2x-b_1y+b_2x&=-c_2 \end{align*} $$


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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.