Why the equation of an arbitrary straight line in complex plane is $zz_o + \bar z \bar z_0 = D$ Why the equation of an arbitrary straight line in complex plane is $zz_o + \bar z \bar z_0 = D$ where D $\in R$
I understand that a vertical straight line can be defined by the equation $z+\bar z= D$ because suppose $z =x+yi$ then $\bar z = x-yi$   Thus, $z+\bar z = x+yi+x-yi=2x$ which is an arbitrary vertical straight line in w-plane.
But why $zz_o + \bar z \bar z_0 = D$ is an arbitrary straight line in complex plane?
 A: You know that a vertical straight line can be defined as $z+\bar z= D$, so if you rotate it's points with angle $\theta$ you get $(e^{i\theta}z)+ \overline{(e^{i\theta}z)}= D$ or $e^{i\theta}z + e^{-i\theta}\overline{z}= D$ and with arbitrary real $r\neq0$,
$$re^{i\theta}z + re^{-i\theta}\overline{z}= rD$$
gives us
$$zz_0 +\bar{z}\bar{z_0}= D_0$$
where $z_0=re^{i\theta}$ and $D_0=rD$.
A: HINT:
$$z+\bar z=2\text{Re}(z)\implies zz_0+\bar z\bar z_0=2\text{Re}(zz_0)$$
A: Hint: given any two points $z_1, z_2 \in \mathbb{C}\,$, then $z$ is collinear with $z_1, z_2$ iff there exists $\lambda \in \mathbb{R}$ such that $z-z_1 = \lambda(z-z_2)$. Eliminate $\lambda$ between the following, then define $z_0, D$ appropriately:
$$
\begin{cases}
\begin{align}
z-z_1 &= \lambda(z-z_2) \\
\bar z- \bar z_1 &= \lambda(\bar z- \bar z_2)
\end{align}
\end{cases}
$$
A: Hack the equation.
Substitute:
$$
\begin{cases}
z = x + iy   \\
z_0 = x_0 + iy_0 
\end{cases}
$$
Do some algebraic manipulations and you'll obtain the equation of a line.
Maybe what confuses you is that neither $z_0$ nor $D$ have a clear geometric interpretation in terms of intercepts, distance to the origin, etc...
Hope this helps!
