Equation of a line in complex plane. Lets say I want to find the equation of line passing through $z, z^\prime \in \Bbb C$. What I will do is to solve $\Im (z) = m\Re (z) + c$ and $\Im(z^\prime) = m \Re(z^\prime) + c$ for $m$ and $c$, but I want to solve directly into $zz_0 + \bar z \bar z_0 = c$ for $z_0$ and $c$. Is this possible ? If so how can I do it ?
 A: There's really no difference in what you are trying to do between the complex plane and the $x$-$y$ plane
Let $z = \Re(z) + i \Im(z) = x_1 + iy_1  = (x_1, y_1)$
Let $z' = \Re(z') + i \Im(z') = x_2 + iy_2  = (x_2, y_2)$
The equation for the line is $(y - y_1) = \frac {y_2 - y_1}{x_2 - x_1} (x - x_1)$
Or $y = mx + b$ where $m = \frac {y_2 - y_1}{x_2 - x_1}$ (the slope) and $b = y_1 - mx_1$ (the $y$ intercept).
So the equation for the line is simple $\frac {\Im(w) - \Im(z)}{\Re(w) - \Re(z)} = \frac {\Im(z') - \Im(z)}{\Re(z') - \Re(z)}$
(or any other equivalent way of writing it).

Perhaps a more "mature" way to do it is to express in term of a variable $t$ where $f(0) = z$ and $f(1) = z'$ then $f(t) = t(z'-z)+ z$.
Idea being:
$$\begin{align*}f(t) &= t \cdot (\text{distance traveled in a time unit}) + \text{starting point} \\\\
&= t \cdot (\text{ending point} - \text{starting point}) + \text{starting point} \\\\
&= t \cdot (z' - z) + z. \end{align*}$$
A: You would need to express $z$ parametrically. For example, to find a line that passes thru $z_1,z_2$ you could say
$$z=z_1+(z_2-z_1)~t$$
When $t=0, z=z_1$ and when $t=1, z=z_2$. And of course, you can have $t\lt0$ and $t
\gt1$. $t$ doesn't even have to be linear, for example, $z=z_1+(z_2-z_1)~\cos t$ will also give a straight line.
