How is this an equation of a line? First let me start that I have read some related questions on MathStackExchange, yet none really answers my question.
I am trying to understand the proof that Möbius transformations map lines and circles into lines and circles. But I am stuck already at the beginning, as I can not, no matter what I do, understand the complex equation of a line.
We wrote than a general equation of a line in $\mathbb{C}$ is: $\text{Re}(\overline{\alpha}z) = b$. How is that a line? How to see it geometrically and understand that this is a line? Some of the things I have written that the professor said are:

*

*This is actually a scalar product.

*When we multiply by $i$, we rotate by $90 ^{\circ}$, so the points that are perpendicular on $\alpha$, are exactly of the form $i \alpha$, so they are in the kernel, then we also have a translation for $b$.

These comments make no sense (to me), as I cannot see in any way, how this is even an equation of a line. I would appreciate if someone thoroughly explains how this represents a line and the direct geometric intrepretation of it.
 A: First, notice that
\begin{equation*}
\mathrm{Re}(z) = b
\end{equation*}
is the "vertical" line that passes through $(b,0)$.
Also, notice that you can divide everything by $|\alpha|$, and assume that $|\alpha| = 1$:
\begin{equation*}
  \mathrm{Re}\left(\frac{\alpha}{|\alpha|}z\right) = \frac{b}{|\alpha|}.
\end{equation*}
Given that $|\alpha| = 1$ and identifying $\mathbb{C}$ and $\mathbb{R}^2$, then the transform
\begin{align*}
  T: \mathbb{R}^2 &\rightarrow \mathbb{R}^2
  \\
  z &\mapsto \alpha z
\end{align*}
is (real) linear. It is actually a rotation. And the inverse rotation is given by the complex conjugate $\overline{\alpha}$.
Every line is the rotation (by some $\alpha$) of a vertical line passing through some $(b,0)$. In other words, for every line, you can arrange $\alpha \in \mathbb{C}$ and $b \in \mathbb{R}$ such that when you apply (the rotation) $\overline{\alpha}$ to it, you get the vertical line through $(b,0)$. And of course, if you rotate a set and get a "vertical line", this set ought to be a line in the first place.
That is, every line (in $\mathbb{R}^2$) is of the form
\begin{equation*}
\mathrm{Re}(\overline{\alpha} z) = b,
\end{equation*}
and every set of this form is a line. In words, if you rotate it back, then it becomes a vertical line.

About you professor comments:

*

*Let $\alpha = (a,b)$ and $z = (x,y)$.
The operator
\begin{align*}
  p: \mathbb{R}^2 \times \mathbb{R}^2 &\rightarrow \mathbb{R}
  \\
  (\alpha, z) &\mapsto \mathrm{Re}(\overline{\alpha}z) = ax + by
\end{align*}
is the so called scalar product. If you understand the "equation of the line" as
\begin{equation*}
  ax + by = c,
\end{equation*}
then this is just $\mathrm{Re}(\overline{\alpha}z) = c$.


*Multiplying by the complex $i$ is just equal to
\begin{align*}
  R: \mathbb{R}^2 &\rightarrow \mathbb{R}^2
  \\
  (x,y) &\mapsto (-y,x).
\end{align*}
This is just the same as sending the $x$-axis to the $y$-axis, and sending the $y$-axis to minus the $x$-axis. That is, you are rotating 90 degrees counterclockwise.
A: First of all, you can check that $\{z\in\mathbb C|\mathrm{Re}(\alpha z)= b\}$ is really an equation of a line if you simply write $z=x+yi$ and $\alpha = c+di$. In that case, you have
$$\begin{align}
\mathrm{Re}(\alpha z) &= \mathrm{Re}((c+di)(x+yi))\\
&=\mathrm{Re}(cx + cyi + dxi - dy)\\
&= \mathrm{Re}((cx-dy) + (cy+dx)i)\\
&= cx-dy
\end{align}$$
which means that
$$\{z\in\mathbb C|\mathrm{Re}(\alpha z)= b\} = \{y\in\mathbb C|z=x+iy, cx-dy = b\}$$
which is clearly a line.

As for your professor's comments are concerned, I can only guess at what is in his head, but this would be my educated guess:

This is actually a scalar product.

He probably means that the calculation of the real part of a number is just a scalar product. This is true. If you view $\mathbb C$ as a 2D vector space over $\mathbb R$, then the mapping $z\mapsto \mathrm{Re}(z)$ is equivalent to the mapping that takes $z$ and calculates the scalar product of $z$ and $1$.

When we multiply by $i$, we rotate by $90 ^{\circ}$, so the points that are perpendicular on $\alpha$, are exactly of the form $i \alpha$, so they are in the kernel, then we also have a translation for $b$.

Here, he is probably talking about the mapping $z\mapsto \mathrm{Re}(\alpha z)$ which is a linear map. As such, it has a kernel. In general, if you have a linear map $\mathcal L: U\to V$, and $v$ is some element of $V$, then the set of solutions to the equaition $\mathcal Lx = v$ will be of the form $\{u_0 + u| u\in \ker(\mathcal L)\}$
A: Let $a=p+qi$ and $z=x+yi$
Thus $\bar a z=(p-qi)(x+yi)=px+qy+pyi-qxi$
So $\Re(\bar a z)=px+qy=b$
Simplifying the factors yields: $x+y\frac{q}{p}=c$
So $\Re(\bar a z)=b$  describes a line with the slope of $a$ in the complex plane.
