Hermitian matrices as circles and lines

To complete our logic we ought to show that an equation such as $$0 = \begin{bmatrix}\bar{z} & 1\end{bmatrix} \begin{bmatrix}a... ...\\ \bar{b} & c \end{bmatrix}\begin{bmatrix}z\\ 1\end{bmatrix},$$ where $H=\begin{bmatrix}a & b \\ \bar{b} & c \end{bmatrix}$ is a Hermitian matrix of negative determinant, always determines a circle or a straight line.

We simply have to perform the algebra in reverse. When you multiply it out, the equation is $$a z\bar{z} +\bar{b}z+b \bar{z}+c = 0.$$ The determinant is $ac-b\bar{b} =ac -\vert b\vert^2$.

If $a\neq 0$, we can divide by $a$ and obtain $$z\bar{z} +\frac{\bar{b}}{a} z+\frac{b}{a} \bar{z} +\frac{c}{a}=0.$$ We can complete the square to put this in the form $$(z+\frac{b}{a})\bar{(z+\frac{b}{a})} = \left\vert\frac{b}{a}\right\vert^2-\frac{c}{a}$$ or $$\left\vert z+\frac{b}{a}\right\vert^2= \frac{\vert b\vert^2-ac}{a^2}.$$ Since the determinant is negative, the right hand side is positive and we see this is a circle with
$$\begin{align*} \mathbf{CENTER}&= -\frac{b}{a} & \mathbf{RADIUS}&= \frac{\sqrt{\vert b\vert^2-ac}}{\vert a\vert} \end{align*}$$

If $a=0$, then since the determinant is negative, we must have $b\neq 0$. Then the equation becomes $$\bar{b}z+b \bar{z}+c = 0$$ which is the equation of a straight line. One point that belongs to this line is $-c/2\bar{b}$. The direction vector of this line is $ib$.

This completes the proof that Hermitian matrices of negative determinant correspond in a one-to-one way with the set of circles and straight lines in the complex plane.