Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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

2 Answers 2

up vote 3 down vote accepted


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.

share|improve this answer
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.

share|improve this answer
Thanks you alex.jordan. –  Dedede Feb 8 '12 at 3:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.