# The inversion of a line/circle is a line/circle in the complex plane

By now I know that a line or a circle in the complex plane can be described as the set of solutions of the equation where $$b$$ is complex and $$a,c$$ is real

$$a|z|^2+\bar{b}z+b\bar{z}+c=0$$

That is because if I take a circle the it is set of all Elements which can be described as the solutionset of the familiar equation $$|z-z_0|=r$$ which is equivalent to $$|z-z_0|^2=r^2$$ which is equivalent to $$(z-z_0)\overline{z-z_0}-r^2=0\iff |z|^2-\bar{z_0}z-z_o\bar{z}+|z_0|^2-r^2=0$$ Then $$a=1,b=z_0,c=|z_0|^2-r^2$$ and I get the desired equation.

Also I can describe a line in this way. Because a line in $$\mathbb{R}^2$$ can be described as $$v+\mathbb{R}w$$ where $$v,w\in\mathbb{R}^2$$ and $$w\neq 0$$. One can prove that such a line is the set of solutions of a equation $$qx+ry+m=0$$ and conversely each solutionset of such an equation can be described as a line in $$\mathbb{R}^2$$. So I take an equation $$qx+ry+m=0$$, and I want to express it as $$a|z|^2+\bar{b}z+b\bar{z}+c=0$$.$$a|z|^2+\bar{b}z+b\bar{z}+c=0$$ can also be expressed as $$ax^2+ay^2+2(b_1x+b_2y)+c=0$$. Where $$z=(x,y),b=(b_1,b_2)$$. Now by setting $$\frac{q}{2}=b_1$$ and $$\frac{r}{2}=b_2$$,i.e $$b=(\frac{q}{2},\frac{r}{2})$$ and $$a=0$$ and $$c=m$$ we have desired equation.

Conversely if we look at the solutionset of an equation $$a|z|^2+\bar{b}z+b\bar{z}+c=0$$ and assume $$a$$ is not equal to $$0$$ then seeing the pattern $$ax^2+bx+cx$$ should be the cue for completing the square. In this case we note that $$z\bar{z}+z\bar{\frac{b}{\bar{a}}}+\bar{z}\frac{b}{a}+\frac{b^2}{a^2}$$. In the second sumand I wrote a bar over the $$a$$ but since $$a$$ is a real number $$\bar{a}=a$$ (then $$a^2=|a|^2$$).

Also $$z\bar{z}+z\bar{\frac{b}{\bar{a}}}+\bar{z}\frac{b}{a}+\frac{|b|^2}{|a|^2}=(z+b/a)\overline{(z+b/a)}=|z+b/a|^2$$

Now $$a|z|^2+\bar{b}z+b\bar{z}+c=0\iff |z|^2+\frac{\bar{b}}{a}z+\frac{b}{a}\bar{z}+\frac{c}{a}+\frac{|b|^2}{a^2}-\frac{|b|^2}{a^2}=0\iff |z+b/a|^2=(|b|^2-ac)/a^2\iff |z+b/a| = \sqrt{(|b|^2-ac)/a^2}$$

Setting $$z_0 = -(b/a)$$ and $$r=\sqrt{(|b|^2-ac)/a^2}$$ we get the equation $$|z-z_0|=r$$ whose solutionset describes a circle.

If a is not equal to Zero then we have $$\bar{b}z+b\bar{z}+c=0$$ One can rewrite this again as $$2(b_1x+b_2y)+c=0$$ Then $$2b_1=q,2b_2=r$$ and $$c=m$$. We again get an equation $$qx+ry+m=0$$ where the solution set describes this time a line.

## $$\text{Question}$$

Let $$f:\mathbb{C}^*\rightarrow\mathbb{C}^*$$ and $$f(z)=\frac{1}{z}$$. Now I take a circle $$C$$ and the Point $$(0,0)$$ does not lay in the circle. The circle can be described as $$\{z\in\mathbb{C}:|z^2|+\bar{b}z+b\bar{z}+c=0\}$$ Why is $$f(C)=\{z\in\mathbb{C}:1+\bar{b}\frac{1}{\bar{z}}+b\frac{1}{z}+c|\frac{1}{z}|^2=0\}$$ (There is a bar over $$z$$ in the second summand)?

I don't understand why $$f(C)$$ must be a circle again. I am confused because the variable has a new name. Geometrically I can draw a picture but just from the equations I cannot see it. I also don't understand how one came up with the equation. It says that if I pick a $$z_0$$ which fullfils the condition $$A:|z_0|^2+\bar{b}z_0+b\bar{z_0}+c=0$$ then $$\frac{1}{z_0}$$ must fulfil the condition $$B:1+\bar{b}\frac{1}{\bar{z_0}}+b\frac{1}{z_0}+c|\frac{1}{z_0}|^2$$.

I can get $$1+\bar{b}\frac{1}{\bar{z_0}}+b\frac{1}{z_0}+c|\frac{1}{z_0}|^2$$ by multiplying $$|z_0|^2+\bar{b}z_0+b\bar{z_0}+c=0$$ with $$\frac{1}{|z_0|^2}$$. But why does $$\frac{1}{z_0}$$ now fullfils the condition $$B$$?

To phrase it differently how can I show $$f(C)=\{z\in\mathbb{C}:1+\bar{b}\frac{1}{\bar{z}}+b\frac{1}{z}+c|\frac{1}{z}|^2=0\}$$ properly and why is $$f(C)$$ then a circle?

The other Questions are similar if $$C$$ is a circle that intersects $$(0,0)$$ then $$C=\{z\in\mathbb{C}:|z|^2+\bar{b}z+b\bar{z}=0\}$$. I know this it follows from $$|z+b/a|^2=(|b|^2-ac)/a^2$$. What I don't understand is that why $$f(C)=\{z\in\mathbb{C}:1+\bar{b}\frac{1}{\bar{z}}+b\frac{1}{z}=0\}$$ and why is it a line which does not go through $$(0,0)$$

If $$L$$ is a line that does not go through $$(0,0)$$ then because for some line which is expressed as the solutionset of $$qx+ry+m=0$$ the corrsponding parametric Version would be $$(\frac{-m}{q},0)+\mathbb{R}(-\frac{r}{q},1)$$ and because $$m=c$$ (and $$a=0$$) for the corresponding equation $$a|z|^2+\bar{b}z+b\bar{z}+c=0,\, c\neq 0$$. So $$L$$ can be expressed as (solutionset) of $$\bar{b}z+b\bar{z}+c=0$$. Why is $$f(C)=\{z\in\mathbb{C}:\bar{b}\frac{1}{\bar{z}}+b\frac{1}{z}+c|\frac{1}{z}|^2=0\}$$ and why is $$f(C)$$ then a circle that does not go through Zero?

The last case is a line $$L$$ that intersects $$(0,0)$$ witht the same Argumentation as above one can show that $$L$$ can be expressed as (a solutionset of) $$\bar{b}z+b\bar{z}=0$$, i.e c must be $$0$$. Why is $$f(L)=\{z\in\mathbb{C}:\bar{b}\frac{1}{\bar{z}}+b\frac{1}{z}=0\}$$ and why is it then a line that 'goes through' Zero but does not contain $$0$$?

Also what happens in the cases where the circle or line intersects the origin $$(0,0)$$ to the values which get close to Zero?

By showing properly I mean that I first take an element of the first set and then show that it is in the other set. The other Question: Why is this a circle or a line? How can I patternmatch the set in Question with an identity of a circle/line that I have proved earlier.

I hope somebody can help.

It's a long question, so maybe I missed something. First of all, you found out in order to be a circle, the coefficient of $$|z|^2$$ must be non zero. Divide your original equation by $$|z|^2=z\bar z$$. You can do that, since $$z\ne 0$$ (the $$(0,0)$$ point is not on the original circle). What you get is $$1+\bar b\frac z{z\bar z}+b\frac{\bar z}{z\bar z}+c\frac 1{|z|^2}=0$$ If you call $$w=\frac 1z$$, you have $$\bar w=\overline{\left(\frac 1z\right)}=\frac 1{\bar z}$$ Then the equation for $$w$$ is $$c|w|^2+\bar b w+b\bar w+1=0$$ This is an equation for a circle, if and only is $$c\ne 0$$. So how can we prove that? Plug in $$z=0$$ in the first equation. You know that $$z=0$$ is not on the circle, so $$|0|^2+\bar b 0+b0+c\ne 0$$, or $$c\ne 0$$. Therefore, the equation for $$w=\frac 1z$$ is a circle.

Different circles are parameterized by different values of $$a$$, $$b$$, and $$c$$.

If your original circle is given by the set of $$z$$ for which

$$a_1 |z|^2 + \bar{b}_1 z + b_1 \bar{z} + c_1 = 0$$

then the inverted circle is given by the set of $$z$$ for which

$$a_2 |z|^2 + \bar{b}_2 z + b_2 \bar{z} + c_2 = 0$$

where $$a_2 = c_1$$ and $$b_2 = \bar b_1$$, and $$c_2 = a_1$$.

Note that the formula corresponds to a circle only when the radius is positive, so $$r^2 = \frac{b \bar b}{a^2} - \frac{c}{a} > 0$$.
We can think of a line as a special case of the circle formula, namely when $$a=0$$ and $$b \ne 0$$.
And note that for both circles and lines, the shape includes the origin if and only if $$c=0$$.

Derivation

You can derive this with the approach you were using. You just have to be a bit more careful about which values of $$(a,b,c)$$ go with which shape.

The original circle or line is given by the set of points $$z$$ that satisfy

$$a_1 |z|^2 + \bar{b}_1 z + b_1 \bar{z} + c_1 = 0 .$$

We want to consider the set of points $$z$$ which have the property that their inverse, $$1/z$$, is on the original circle or line. Whatever the overall shape of this new set may be, we know it is the set of points satisfying

$$a_1 \left|\left(\frac 1z\right)\right|^2 + \bar{b}_1 \left( \frac 1z \right) + b_1 \overline{\left( \frac 1z \right)} + c_1 = 0$$

$$a_1 \frac{1}{|z|^2} + \bar{b}_1 \frac 1z + b_1 \frac{1}{\bar z} + c_1 = 0$$

Multiplying both sides by $$|z|^2$$, which is the same as $$z \bar z$$, we get

$$a_1 + \bar{b}_1 \bar z + b_1 z + c_1 |z|^2 = 0$$

$$c_1 |z|^2 + b_1 z + \bar{b}_1 \bar z + a_1 = 0$$

If we define $$a_2 = c_1$$ and $$b_2 = \bar b_1$$, and $$c_2 = a_1$$, then we can write our equation as:

$$a_2 |z|^2 + \bar b_2 z + b_2 \bar z + c_2 = 0$$

And this is exactly the form of the equation for a circle, since $$a_2$$ and $$c_2$$ are real.
(Or a line, if $$a_2=0$$, which happens if $$c_1=0$$, which means the original circle/line went through the origin.)

In order to be sure that the new shape is really a circle or line, we still need to check the condition that $$\frac{b \bar b}{a^2} - \frac{c}{a} > 0$$ (the case of a circle) or else $$a=0$$ and $$b \ne 0$$ (the case of a line). We can write this as

$$\frac{b_2 \bar b_2}{a_2^2} - \frac{c_2}{a_2} > 0 \quad \mbox{or} \quad a_2 = 0 \mbox{ and } b_2 \ne 0$$

$$\frac{b_1 \bar b_1}{c_1^2} - \frac{a_1}{c_1} > 0 \quad \mbox{or} \quad c_1 = 0 \mbox{ and } b_1 \ne 0$$

We know that we cannot have $$c_1 = b_1 = 0$$, since that corresponds neither to a circle nor to a line for the original shape. Now either $$c_1 = 0$$ and $$b_1 \ne 0$$, and the second condition holds, or else $$c_1 \ne 0$$, in which case the second condition does not hold, and we must check the first condition (the inequality). In this case, $$c_1^2$$ is strictly positive, and so we can multiply both sides of the inequality by $$c_1^2$$ to get

$$b_1 \bar b_1 - a_1 c_1 > 0 .$$

So, is this condition satisfied?

Well, if the original shape was a line, then $$a_1 = 0$$ and $$b_1 \ne 0$$, and the condition is satisfied.
On the other hand, if the original shape was a circle, then we know that $$a_1 \ne 0$$ and $$\frac{b_1 \bar b_1}{a_1^2} - \frac{c_1}{a_1} > 0 .$$

Again, since $$a_1^2 > 0$$, we can multiply both sides by $$a_1^2$$ to get $$b_1 \bar b_1 - c_1 a_1 > 0 ,$$ and so the condition is satisfied in this case as well.

Thus we see that when we invert a circle or line, we not only get a set of points satisfying the "circle equation", but the parameters of the equation also satisfy the criteria for actually corresponding to a circle or line.