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

Question

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

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

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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.

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$$\text{Question}$$

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.

Answer 1

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.

Answer 2

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 $c=0$.

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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.