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.
 A: 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.
A: 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.
A: 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
$$w=\frac{(w_1-\alpha w_3)z+(w_1+\alpha w_3)}{(1-\alpha)z+(1+\alpha)}.$$
A: It is more straight about this answer.
For $z_1,z_2,z_3\in\mathbb{R}$, LFT $f$ such that $f(z_i)=\omega_i\in \mathbb{R},i=1,2,3$, is unique. Cross ratio
      $$
  \frac{\omega-\omega_2}{\omega-\omega_3}:
  \frac{\omega_1-\omega_2}{\omega_1-\omega_3}=
  \frac{z-z_2}{z-z_3}:\frac{z_1-z_2}{z_1-z_3}.
   $$
        Simplify the last equation, we have 
        \begin{align*}
     \omega&=\frac{(z-z_2)(z_1-z_3)(\omega_1-\omega_2)\omega_3-(z-z_3)(z_1-z_2)(\omega_1-\omega_3)}{(z-z_2)(z_1-z_3)(\omega_1-\omega_2)-(z-z_3)(z_1-z_2)(\omega_1-\omega_3)\omega_2}\\
   &=\frac{[(\omega_1-\omega_2)(z_1-z_3)\omega_3-(z_1-z_2)(\omega_1-\omega_3)]z-[z_2\omega_3(\omega_1-\omega_2)(z_1-z_3)+z_3(z_1-z_2)(\omega_1-\omega_3)]}{[(\omega_1-\omega_2)(z_1-z_3)-(z_1-z_2)(\omega_1-\omega_3)\omega_2]z-[z_2(\omega_1-\omega_2)(z_1-z_3)+z_2\omega_3(z_1-z_2)(\omega_1-\omega_3)]}.
  \end{align*}
It follows that all LFTs maping real axis to itself have the form $$\omega=\displaystyle\frac{az+b}{cz+d},a,b,c,d\in \mathbb{R}, ad-bc\neq 0.$$
Next, we prove for all forms like $\dfrac{az+b}{cz+d},a,b,c,d\in\mathbb{R},ad-bc\neq 0$ map real axis to real axis.
Notice that
      $$
  \mathrm{Im}\ \omega=\frac{1}{2i}(\omega-\overline{\omega})=\frac{ad-bc}{|cz+d|^2}\mathrm{Im}\ z.
   $$
        That is, $\mathrm{Im}\ \omega=0$ if and only if $\mathrm{Im}\ z=0$.
A: Let $w=\frac{az+b}{cz+d}$ where $ad-bc\neq0$. Suppose that $a=a_1+ia_2,~b=b_1+ib_2,~c=c_1+ic_2,~d=d_1+id_2$ and $w=u+iv$, where $a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2,u,v\in\mathbb R$.
Then $z=\frac{dw-b}{-cw+a}$ and after some manipulation we see that $z=\frac{A+iB}{C^2}$ where
$$B=(d_1c_2-d_2c_1)(u^2+v^2)+(a_1d_2-a_2d_1+b_2c_1-b_1c_2)u+(a_1d_1-b_1c_1+a_2d_2-b_2c_2)v+(b_1a_2-a_1b_2)$$
Since $\Im z=0$ implies that $\Im w=0$, we conclude that
$$d_1c_2-d_2c_1=0\tag{1}$$
$$a_1d_2-a_2d_1+b_2c_1-b_1c_2=0\tag{2}$$
$$b_1a_2-a_1b_2=0\tag{3}$$
It follows from (1) and (3) that $a_2=ma_1,~b_2=mb_1,c_2=nc_1$ and $d_2=nd_1$ for some constants $m,n\in\mathbb R$.
Since $ad-bc=(1-mn)(a_1d_1-b_1c_1)+i(m+n)(a_1d_1-b_1c_1)\neq0$ we conclude that $a_1d_1-b_1c_1\neq0$. It follows from (2) that
$$na_1d_1-ma_1d_1+mb_1c_1-nb_1c_1=(n-m)(a_1d_1-b_1c_1)=0$$
which implies that $m=n$. Therefore,
$$w=\frac{(a_1+ima_1)z+b_1+imb_1}{(c_1+imc_1)z+d_1+imd_1}=\frac{(1+im)a_1z+(1+im)b_1}{(1+im)c_1z+(1+im)d_1}=\frac{a_1z+b_1}{c_1z+d_1}.$$
