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

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.

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I'm sorry, but are you asking why a fractional linear transformation (from $\mathbb{C}\cup\{\infty\}$ to itself) which maps $\mathbb{R}$ to itself must have real coefficients, or why a linear (and if so, linear over what? $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$? from what space to what space?) transformation maps the real line to the real line? – Arturo Magidin Feb 5 '12 at 2:48
The conclusion certainly does not follow: take $a=d=i$, $b=c=0$. The map is the identity, but the coefficients are not real. – Arturo Magidin Feb 5 '12 at 2:51
@ArturoMagidin I'm sorry for being confusing. The exact statement is Show that any linear transformation which transforms the real axis into itself can be written with real coefficients. I assumed it meant linear fractional (from $\mathbb{C}\cup\{\infty\}$ to itself), since those were the only transformations discussed in the book, and I assume it's linear over $\mathbb{C}$. – Dedede Feb 5 '12 at 2:53
If a fractional linear transformation is linear, then it maps $0$ to $0$, so $b=0$. It also maps $\alpha$ to $\alpha f(1)$, hence $a(c+d) = a(c\alpha+d)$ for all $\alpha$; thus, either $a=0$, or $c=0$. If $a\neq 0$, then it is just multiplication by a scalar, $\frac{a}{d}$, which must be real (since it maps $1$ to a real); so you can write it with real coefficients. If $a=0$, then its the zero map, and you can write it with real coefficients. – Arturo Magidin Feb 5 '12 at 2:55
A linear fractional transformation maps $\mathbb R\cup\{\infty\}$ to itself if and only if it can be written with real coefficients. Given the problem as stated, $\infty$ is sent to $\infty$, and Arturo has shown that the map has the form $z\mapsto \alpha z+\beta$ with $\alpha$ and $\beta$ in $\mathbb R$. Otherwise, if $c\neq 0$ you can divide all coefficients by $c$, then show that if $z\mapsto \frac{a'z+b'}{z+d'}$ maps $\mathbb R\cup\{\infty\}$ to $\mathbb R\cup\{\infty\}$ then $a'$, $b'$ and $d'$ are real. (The easiest part is noting that $\infty$ is sent to $a'$, so $a'$ is real.) – Jonas Meyer Feb 5 '12 at 4:20

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.

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Doesn't it seem like the OP will also be considering transformations that take $\mathbb{R}\cup\{\infty\}$ to itself? – alex.jordan Feb 5 '12 at 4:10
@alex.jordan: Hmm... Worse; $\frac{d}{c}$ need not be real at the point I am discussing it. – Arturo Magidin Feb 5 '12 at 4:14
Thank you for the detailed answer! – Dedede Feb 8 '12 at 3:59

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.

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Thanks you alex.jordan. – Dedede Feb 8 '12 at 3:59

If $w=f(z)$ is a linear fractional transformation that transforms the real axis into itself, setting $z_1=1, z_2=0, z_3=-1$, we know $w_1=f(z_1), w_2=f(z_2), w_3=f(z_3)$ are all real. Using the cross ratio, we have

$$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\frac{-(z-1)}{z+1}.$$

Note $\frac{w_2-w_3}{w_2-w_1}$ is real, so we can rewrite

$$\frac{w-w_1}{w-w_3}=\alpha\frac{z-1}{z+1}$$

where $\alpha$ is some real number. Solving for $w$, we have

$$w=\frac{(w_1-\alpha w_3)z+(w_1+\alpha w_3)}{(1-\alpha)z+(1+\alpha)}.$$

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