Any linear fractional transformation transforming the real axis to itself can be written in terms of reals?

Question

I'm trying to teach myself complex analysis, and was reading about linear transformations.

I would like to understand why any linear fractional transformation which transforms the real axis into itself can be written with real coefficients.

I assume I have some linear fractional transformation $z\mapsto \frac{az+b}{cz+d}$, where $z\in\mathbb{C}$ and $a,b,c,d\in\mathbb{C}$ are the coefficients, and I think I would like to conclude $a,b,c,d\in\mathbb{R}$ actually.

Choosing various reals, I get $$ 0\mapsto\frac{b}{d},\quad 1\mapsto\frac{a+b}{c+d},\quad -1\mapsto\frac{-a+b}{-c+d} $$ so I know all those images are again real. Is there someway to conclude that $a,b,c,d$ are individually real? Thank you.

Answer 1

Edited.

I'm assuming we want $\mathbb{R}\cup\{\infty\}$ to be mapped to $\mathbb{R}\cup\{\infty\}$. The desired conclusion is that we can find $a',b',c',d'\in\mathbb{R}$ such that $$\frac{az+b}{cz+d} = \frac{a'z+b'}{c'z+d'}\quad\text{for all }z\in\mathbb{C}.$$

By plugging in $0$, we conclude that either $d=0$ or $\frac{b}{d}\in\mathbb{R}$.

If $d=0$, plugging in $\infty$ gives $\frac{a}{c}$, hence $\frac{a}{c}=\alpha\in\mathbb{R}$ (or $c=0$, in which case the transformation just gives $z\mapsto \infty$ for all $z$, and we can rewrite it as $\frac{1}{0z+0}$). We can rewrite the transformation as: $$\frac{az+b}{cz} = \alpha + \frac{b}{cz}.$$ Plugging in $z=1$ gives $\alpha+\frac{b}{c}\in\mathbb{R}$, hence $\frac{b}{c}=\beta\in\mathbb{R}$; so we can rewrite $$\frac{az+b}{cz+d} = \frac{az+b}{cz} = \alpha + \frac{\beta}{z} = \frac{\alpha z+\beta}{1z+0},$$ and we are done.

If $b=0$, then composing with $z\mapsto \frac{1}{z}$ we can repeat the argument above. So we may assume that $d\neq 0$ and $b\neq 0$.

Then $\frac{b}{d}=\beta\in\mathbb{R}$, so we can rewrite as $$ \frac{az+b}{cz+d} = \frac{az+\beta d}{cz+d},\quad\beta\in\mathbb{R}.$$ Plugging in $\infty$ we get $\frac{a}{c}=\alpha\in\mathbb{R}$, so we can write $$ \frac{az+b}{cz+d} = \frac{\alpha cz + \beta d}{cz+d}.$$ Plugging in $1$ we get $$\frac{\alpha c + \beta d}{c+d} = \alpha + \frac{(\beta-\alpha)d}{c+d}.$$ Since $\alpha$ and $\beta$ are real, this is a real number (or $\infty$) if and only if $\frac{d}{c+d}$ is a real number (or $\infty$), if and only if $\frac{c+d}{d} = \frac{c}{d}+1$ is a real (or $\infty$), if and only if $\frac{c}{d}$ is a real number (cannot be $\infty$, since $d\neq 0$).

Thus, $c=\gamma d$ with $\gamma\in\mathbb{R}$, so $$ \frac{az+b}{cz+d} = \frac{\alpha cz + \beta d}{cz+d} = \frac{\alpha\gamma dz + \beta d}{\gamma dz + d} = \frac{\alpha\gamma z+ \beta}{\gamma z + 1},$$ with $\alpha,\beta,\gamma\in\mathbb{R}$, as desired.

Answer 2

Given $\frac{az+b}{cz+d}$, you can trivially assume that $a$ is real. Either $a=0$, or you can multiply all four numbers by $\overline{a}$. (or divide all four by $a$)

Now that $a$ is real, $c$ can be assumed real too. If $a$ were zero, apply the same trick to $c$. ($c$ is not zero, or else the transformation is noninvertible. So multiply all coefficients by $\overline{c}$ or divide them all by $c$.) If $a$ were nonzero, then note that $\infty$ gets mapped to $\frac{a}{c}$, which needs to be in $\mathbb{R}\cup\{\infty\}$. So either $c=0$ or $\frac{a}{c}$ is real. Well, $a$ is already nonzero real, so $c$ is real too.

Now look at $f^{-1}(z)=\frac{dz-b}{-cz+a}$. We already know $c$ can be assumed real, so the same argument shows $d$ can be assumed real.

Finally your observation about the image of $1$ shows that if $a$, $c$, and $d$ are real, then $b$ is real too.