Find a conformal mapping $f$ that maps $\{z \in \mathbb{C}: Re(z) > 0\}$ onto itself that maps $f(1)=2$

Question

Find a conformal mapping $f$ that maps $\{z \in \mathbb{C}: Re(z) > 0\}$ onto itself that maps $f(1)=2$.

So we need to ensure $f(1)=2$ and that the right half plane maps to the right half plane. Well, we can choose a Mobius mapping. Notice that the mapping

$$f(z) = \frac{z+1}{\frac{1}{2}z+\frac{1}{2}}$$

will satisfy $f(1) =2$ but it's not obvious to me that I can check that the right half plane maps onto the right half plane. In fact, my choice of $f$ above is only one of infinitely many choices that satisfy the constraint $a+b = 2(c+d)$. How can I choose a mapping in such a way that the right half plane maps onto itself?

Answer 1

The correct order of matrix multiplication for this is $$ \left( \begin{array}{rr} -i & 0 \\ 0 & 1 \\ \end{array} \right) \left( \begin{array}{rr} a & b \\ c & d \\ \end{array} \right) \left( \begin{array}{rr} i & 0 \\ 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rr} a & -bi \\ ci & d \\ \end{array} \right) $$

Here we demand $a,b,c,d$ real and $ad-bc > 0.$ Your condition that $1$ map to $2$ by $$ \frac{az-bi}{ciz +d} $$ becomes $ a-bi = 2 (ci +d)$ with real variables, so that $a-2d = (b+2c)i $ gives the restrictions..........

$$ \frac{2dz + 2ci}{ciz +d} $$ from matrix $$ \left( \begin{array}{rr} 2d & 2ci \\ ci & d \\ \end{array} \right) $$ of determinant $2(c^2 + d^2)$

Answer 2

There are plenty of Mobius transformations that do this, but perhaps the simplest one is $f(z)=2z=\frac{2z+0}{0z+1}$.