# Equation of a line in a complex plane- why is my equation wrong?

In the book there is an excercise where I have to Show that every line (and circle) can be expressed in the form $$a|z|^2+\bar{b}z+b\bar{z}+c=0$$ where $$a$$ and $$c$$ is a real number and $$b$$ is a complex number. Translated into the $$\mathbb{R}^2$$ Notation it means that every line (or circle) can be described in the form of $$(ax^2+ay^2+2(a'x+b'y)+c,0)$$ where $$z=(x,y),b=(a',b'),a=(a,0)c=(c,0)$$. I did succed to Show that such a formula exists for a circle but my suggested equation for the line does not fit in this Frame.

My suggested formula for a line does not work. First of all a line $$L$$ is a set with the properties $$\{z\in\mathbb{C}:z=ac\}$$ where $$a$$ is a real number and $$c$$ is a fixed complex number.

The formula I provided was $$|c|z-c|z|=0$$

And I have already shown that $$z$$ is a solution if and only if $$z$$ can be expressed in the form $$ac$$ where $$a$$ is a real number again.

But then again if I set $$c=(q,r)$$ and $$z=(x,y)$$

I get when I write my formula in $$\mathbb{R}^2$$ Notation

$$(\sqrt{q^2+r^2}x,\sqrt{q^2+r^2}y)-(\sqrt{x^2+y^2}q,\sqrt{x^2+y^2}r)$$

$$=(\sqrt{q^2+r^2}x-\sqrt{x^2+y^2}q,\sqrt{q^2+r^2}y-\sqrt{x^2+y^2}r)$$

Since the immaginary part is not Zero the equation does not fit in the Frame

$$(ax^2+ay^2+2(a'x+b'y)+c,0)$$

Maybe my Question is unclear so I repeat it again: Is the set of Solutions to the equation $$(\downarrow)$$ where $$c$$ is a fixed complex number and $$z$$ is the variable $$|c|z-c|z|=0$$ a line in the complex plane? If not then can someone give me an example for two Points of such a function which are consideres as Solutions but are not linear Independent to each other? If the set of Solutions i.e the set of all $$z$$ which solve the equation above are actually a line then how can I describe it in the form: $$a|z|^2+\bar{b}z+b\bar{z}+c=0$$ where $$a$$ and $$c$$ is a real number and $$b$$ is a complex number?

• "First of all a line 𝐿 is a set with the properties $\{𝑧\in\mathbb{ℂ}:z=ac\}$ where $a$ is a real number and $c$ is a fixed complex number." Isn't this a line through the origin, not a general line? – saulspatz Feb 27 at 18:34
• Now I see $(0,0)$ is Always a solution – RM777 Feb 27 at 18:40
• What is the significance of the constant $a$? What if $a=0$? – Somos Feb 27 at 21:31
• @Somos I have modified the equation and now it should work for every line but I am still not able to write the equation in the desired form i.e $a|z|^2+\bar{b}z+b\bar{z}+c=0$ can you have a look at it please? The new function I came up with is $x|z-c|=(z-c)|x|\iff x|z-c|-\big{(}(z-c)|x|\big{)}=0$ Ist Solutions are exactly the Points on the line $\{z\in\mathbb{C}:rx+c=z\}$ where $x,c$ are fixed complexnumbers and $r$ is a real number. The proof is derived from the proof that the set of solutions for $x|z|-|x|z=0$ is exactly the line $\{z\in\mathbb{C}:z=rx\}$ – RM777 Feb 27 at 21:59
• @saulspatz How can I rewrite $x|z-c|-\big{(}(z-c)|x|\big{)}=0$ to $a|z|^2+\bar{b}z+b\bar{z}+c'=0$ for some $b'\in\mathbb{C}$ and $a,c\in\mathbb{R}$? – RM777 Feb 27 at 22:02

Don't forget about the "$$= 0$$" part of the equations. The formula $$(ax^2+ay^2+2(a'x+b'y)+c,0)$$ is not an equation; the equivalent equation to $$a\lvert z\rvert^2+\bar{b}z+b\bar{z}+c=0$$ is $$(ax^2+ay^2+2(a'x+b'y)+c,0) = (0,0),$$ or more simply $$ax^2+ay^2+2(a'x+b'y)+c = 0.$$

For the equation $$\lvert c\rvert z-c\lvert z\rvert=0,$$ in your $$\mathbb R^2$$ notation you get the equation

$$\left(\sqrt{q^2+r^2}x-\sqrt{x^2+y^2}q,\sqrt{q^2+r^2}y-\sqrt{x^2+y^2}r\right) = (0,0).$$

That is equivalent to the two simultaneous equations \begin{align} x\sqrt{q^2+r^2} - q\sqrt{x^2+y^2} &= 0,\tag1 \\ y\sqrt{q^2+r^2} - r\sqrt{x^2+y^2} &= 0.\tag2 \end{align}

We have to assume $$c\neq 0,$$ otherwise your equation is true everywhere. So we can divide this into two cases. In the first case, $$r = 0$$ but $$q \neq 0.$$ In that case, Equation $$(2)$$ says that $$y = 0,$$ and therefore Equation $$(1)$$ says that $$x\lvert q\rvert = q \lvert x\rvert.$$ This is true for all non-negative $$x$$ when $$q > 0$$ and for all non-positive $$x$$ when $$q < 0.$$

In the second case, $$r\neq 0.$$In that case, Equation $$(2)$$ says that $$\sqrt{x^2+y^2} = \frac yr \sqrt{q^2+r^2},$$ and plugging that into Equation $$(1)$$ we get $$x\sqrt{q^2+r^2} - y \frac qr \sqrt{q^2+r^2} = 0.$$ This is an equation of the form $$ax^2+ay^2+2(a'x+b'y)+c = 0$$ where $$a = 0,$$ $$a'= \frac12\sqrt{q^2+r^2},$$ $$b' = -\frac q{2r} \sqrt{q^2+r^2},$$ and $$c = 0.$$ And it would be an equation of a line through the origin, except that Equation $$(2)$$ also implies that $$y$$ has the same sign as $$r,$$ and therefore you only get half of the line.

Starting over from the beginning, a general equation for an arbitrary line in $$\mathbb R^2$$ (expressed in $$(x,y)$$ coordinates) is $$Ax + By + C = 0, \tag3$$ where the coefficients $$A,$$ $$B,$$ and $$C$$ are whatever they need to be in order to produce an equation of the desired line.

An equation of this form exists for any line at any angle (even parallel to the $$y$$ axis) through any point in the plane. For example:

• To represent the line with equation $$y = 3x + 5,$$ simply set $$A=3,$$ $$B=-1,$$ and $$C=5,$$ and then you have the equivalent equation $$3x + (-1)y + 5 = 0$$, which is in the form $$Ax + By + C = 0.$$

• To represent the line $$y = -7,$$ we can set $$A=0,$$ $$B=1,$$ and $$C=7$$ to get the equivalent equation $$0x + 1y + 7 = 0.$$

• To represent the line $$x = 15,$$ simply set $$A=1,$$ $$B=0,$$ and $$C=-15$$ so that the equation $$Ax + By + C = 0$$ becomes $$1x + 0y + (- 15) = 0.$$

In short, the equation $$Ax + By + C = 0$$ is a very handy one (not just for this problem!) and it is good to make it as familiar as equations such as $$y = mx + k$$ and $$x^2 + y^2 = r^2.$$

Surely it is not hard to see how to put the equation $$ax^2+ay^2+2(a'x+b'y)+c = 0$$ into the form of Equation $$(3).$$

• I have modified the equation and now it should work for every line but I am still not able to write the equation in the desired form i.e $a|z|^2+\bar{b}z+b\bar{z}+c=0$ can you have a look at it please? The new function I came up with is $x|z-c|=(z-c)|x|\iff x|z-c|-\big{(}(z-c)|x|\big{)}=0$ Its Solutions are exactly the Points on the line $\{z\in\mathbb{C}:rx+c=z\}$ where $x,c$ are fixed complexnumbers and $r$ is a real number. The proof is derived from the proof that the set of solutions for $x|z|-|x|z=0$ is exactly the line $\{z\in\mathbb{C}:z=rx\}$ – RM777 Feb 27 at 22:05
• With that modification you can start the half-line at any point in the complex plane, but it is still only half a line. The approach is fundamentally flawed that way. My advice would be to try as hard as you can to forget your equation, and start over. I think it is a dead end and is preventing you from seeing the easy and obvious way to solve the problem. – David K Feb 27 at 23:56
• Okey I will try to do that , you saif $y$ has the same sign as $r$. So Solutions where $y$ has not the same sign as $r$ are not considered. So if I have a line $\{𝑧\in\mathbb{ℂ}:z=ac\}$ and if I have a solution $bc$ it would not be a solution to the equation if the signs of the immaginary parts of $b$ and $c$ don't match. I pick as an example the line $\{i\cdot r\}$. Then $-i$ would have to be on the line but$(-i)-i=-2i$. So you were Right after all – RM777 Feb 28 at 15:37
• I came up with a Formula that does not involve absolute values. It is $\frac{Im(z)-b}{Re(z)}-\frac{r}{q}=0$ where $c=(q,r)$ is some complex number and $b$ is pure immaginary. The idea was that if $z$ is a Point on the line $\{b+tc\}$ where $t$ is a real number then the triangle constructed by the Points $(z,Re(z),b)$ must be similair to the triangle $(0,Re(c),Im(c))$. The Problem is however that it does not work for lines parallel to the immaginary axis. – RM777 Feb 28 at 18:17
• I really don't understand Formula $3$ what is $A$ and $B$ and to 'which' line does it correspond to? You already said that every line can be described that way the way you wrote the Formula implies that one can write a line in Dependance to $A$ and $B$. How is this Dependance expressed? – RM777 Feb 28 at 18:32