I am solving a question from Gareth A. Jones book Complex Functions. The question is:
(i) Show that the circle \begin{equation}\label{1} a z \bar{z}+\bar{b} \bar{z}+b z+c=0 \quad(a, c \in \mathbb{R}) \end{equation} (ii) Deduce that the composition of two inversions $I_{C_{1}}{\circ} I_{C_{2}}$ is hyperbolic if the circles $C_{1}$ and $C_{2}$ do not intersect. Show that the composition of the above two inversions is parabolic if $C_{1}$ and $C_{2}$ touch and elliptic if $C_{1}$ and $C_{2}$ intersect in two points.
To solve (ii), one may assume that $C_2=\mathbb{R},$ and use (i). Here's my solution:
Let $C_2=\mathbb{R}.$ Then, $I_\mathbb{R}$ is given by complex conjugation. Let $C_1$ be the circle given by $$a z \bar{z}+\bar{b} \bar{z}+b z+c=0 \quad(a, c \in \mathbb{R})\quad\dots (1)$$ then, $I_{C_1}$ be given by the mapping $$z\mapsto \frac{-\bar{b}\bar{z}-c}{a\bar{z}+b}.$$ This is the general form of any inversion for the circle with equation (1).
Then, we have that $$A=I_{C_1}\circ I_\mathbb{R}:z\mapsto \frac{-\bar{b}z-c}{az+b}.$$ Now, $A\in PGL(2,\mathbb{C});$ let $B$ be the matrix representation of $A$ in $PSL(2,\mathbb{C}).$ Then, as $det(\lambda A)=\lambda^n det(A),$ we have that $$B=\frac{1}{\sqrt{ac-|b|^2}}A=\frac{1}{\sqrt{ac-|b|^2}}\cdot\begin{bmatrix} -\bar{b} & -c \\ a & b \end{bmatrix}.$$ Now, $A$ is hyperbolic if $tr^2(B)>4,$ i.e., if $$(b-\bar{b})^2=b^2+\bar{b}^2-2|b|^2>4(ac-|b|^2).$$ Simplifying, we see that $A$ is hyperbolic if $$(b+\bar{b})^2=4(\Re(b))^2>4ac,$$ i.e. if $(\Re(b))^2>ac.$
--
This is exactly the opposite of what is needed, that is, what I needed to show was that if $A$ is hyperbolic, then $(\Re(b))^2<ac.$
I don't understand what I'm doing wrong, I believe I have misplaced a minus sign or have done a really terrible error.
Help will be extremely appreciated! If there is any information that is further required, please ask!