Find linear transformation from the strip $0Find the general form of the linear transformation from the strip $0<x<1$ to itself.
Attempt:
The linear transformation has the form $f(z)=az+b$.
We need to find where $a$ and $b$ are.
Let $z=x+yi$ and $A=\{z\in\mathbb C:0<x<1\}$. We want $T(A)=A$.
As $T(A)$ is homeomorphism then $$T(\partial A)=\partial T(A)\\ \implies T(ki\ \cup\ 1+ki)=\partial A=ki\ \cup\ 1+ki$$
case 1:$T(ki\ \cup\ 1+ki)=ki$
case 2: $T(ki\ \cup\ 1+ki)=1+ki$
What does this implies ?
Am I on the right track?
Help please
 A: Let $f(z) = az+b$ be such affine transformation, where $a,b\in \mathbb{C}$. Up to composing with $z\mapsto 1-z$ we may suppose that $f$ leaves invariant both boundary lines $\{Re(z) = 0 \}$ and $\{Re(z) = 1 \}$. 
For $f$ to send $\{Re(z) = 0 \}$ to itself, we must have $a$ real and $b$ purely imaginary. To prove this we only need to see what restrictions that $Re(f(0))=0$ and $Re(f(i))=0$ impose.
$$ 0= Re(f(0)) = Re(b) \Rightarrow b \,{\rm purely\, imaginary}. $$
Using this we have
$$ 0 = Re(f(i)) = Re(ia+b) = Re(ia) = -Im(a) \Rightarrow a\in \mathbb{R}.$$
Next, to leave $\{Re(z) = 1 \}$ we must have $a=1$. To see this we just have to compute the real part of $f(1)$ and force it to be $1$.
$$ 1 = Re(f(1)) = Re(a+b) = a .$$
Therefore  $$ f(z) = z + ci, \, {\rm with }\, c\in \mathbb{R},$$
if $f$ preserves both boundary lines. 
To complete, let $g$ be another such transformation that exchanges the boundary lines, then $g(1-z) = z+ bi$ which means $$g(z) = -z + 1+ci,\, {\rm with }\, c\in \mathbb{R}. $$
Conversely, for any given $c\in \mathbb{R}$ the transformations built as above preserve the region $A$.
