As the title suggests, I need help to understand how the author came up with the equation of a line without knowing exactly two points.
It is well known that if points $a$ and $b$ lie on the unit circle and $a \neq b$, the equation of a line passing these points is given by $$z+ab\bar{z}=a+b.$$
Set-up: Let the points $A$, $B$, $C$, $D$ be represented by the complex numbers $a$, $b$, $c$, $d$ respectively.
Referring to the picture above, the author claims that the equations of parallel lines $AA'$, $BB'$ and $CC'$ are given by
$$z+k\bar{z}=a+k\bar{a},$$ $$z+k\bar{z}=b+k\bar{b},$$ and $$z+k\bar{z}=c+k\bar{c}$$ respectively where $k$ is a suitable complex number such that $|k|=1$.
Then it follows that the points $A'$, $B'$ and $C'$ are given by complex numbers $k\bar{a}$, $k\bar{b}$ and $k\bar{c}$ respectively. [This is clear because I can verify it directly by substituting say A' into the LHS of $z+k\bar{z}=a+k\bar{a}$ and I will easily obtain $a+k\bar{a}$= RHS of the equation.]
My question is how is the author able to determine the equation of line $AA'$ without first having known point $A'$?
My last question is suppose we have an equation of line $l_1:z+ab\bar{z}=a+b$. If I have another line, $l_2:z+ab\bar{z}=k$, does it mean that $l_1$ is parallel to $l_2$ (since the "coefficients" infront of $\bar{z}$ is the same)? It is clear for the case of the cartesian form of line equations $y=mx+c$ but do they work the same way in complex form? Any help would be appreciated!
Hint: $\;z+k\bar{z}=a+k\bar{a}\,$ is the equation of a line through $\,a\,$ since:
the equality holds true for $\,z=a\,$, so it passes through $\,a\,$;
for $\,z \ne a\,$ it can be rewritten as $\dfrac{z-a}{\bar z - \bar a} = -k \;\implies\; 2 \arg(z-a) = \arg(-k)\,$ i.e. the segment between $\,z\,$ and $\,a\,$ has a constant slope.