How is this an equation of a line?

Question

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:

1. This is actually a scalar product.
2. 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.

Answer 1

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.

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About you professor comments:

1. 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 $\mathrm{Re}(\overline{\alpha}z) = c$.

2. 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.

Answer 2

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.

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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 $\{u_0 + u| u\in \ker(\mathcal L)\}$

Answer 3

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.