Find the mobius transformations satisfying each of the following. Write your answers in standard form, as $\frac{az+b}{cz+d}$
(a)1$\rightarrow$0,2$\rightarrow$1,3$\rightarrow$$\infty$ (use the cross ratio)
ans: $z_1$=1,$z_2$=2,$z_3$=3
by definition we have: $f(z)$$=$[z,$z_1$,$z_2$,$z_3$] $=$ $\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$ $=$ $\frac{az+b}{cz+d}$
Then simply plug in values: $f(z)$$=$ $\frac{(z-1)(2-3)}{(z-3)(2-1)}$ $=$ $\frac{-z+1}{z-3}$
(a) is correct and I am having difficulty with (b) and (c)
(b)1 $\rightarrow$0, 1+i $\rightarrow$1, 2 $\rightarrow$ $\infty$ (use the cross-ratio)
I presume $z_1$=1, $z_2$=1+$i$, $z_3$=2
Then i get $\frac{(z-1)(-1+i)}{(z-2)(i)}$. This does match the standard form an seems incorrect.
(c)0$\rightarrow$$i$, 1 $\rightarrow$ 1, $\infty$ $\rightarrow$ -$i$ (does not say to use cross-ratio)
I am presuming we cannot just use cross ratio directly. In my attempt:
0$\curvearrowleft$i, 1$\curvearrowleft$1, $\infty$ $\curvearrowleft$ -$i$ denoted by g
$f(0)$=$i$, $f(1)$=1, $f(\infty)$=-1
$f(z_1)$= 0, $f(z_2)$ = 1, $f(z_3)$ = $\infty$
so
$g(i)$ = 0, $g(1)$ = 1, $g(-i)$ = $\infty$
$w_1$=i, $w_2$=1, $w_3$=-i
$g(w)$ = $\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}$
then z=$\frac{w-i}{w+i}$ $\frac{1+i}{1-i}$
w=f(z)
thus $f(z)$= $\frac{(z-i)(z+i)}{(z+i)(1-i)}$ = $\frac{z-i}{1-i}$
This is incorrect answer.
Best Answer
You have got a right answer on part (b). Just have to write it as $$ \frac{(i-1)z+(1-i)}{iz - 2i}.$$ So it is in standard form.
For the part (c), your $g(w)$ is right. Explicitly $$ g(w) = \frac{(1+i)z + (1-i)}{(1-i)z + (1+i)}. $$ It corresponds to the matrix $$ M = \left[ \begin{array}{cc} 1+i & 1-i \\ 1-i & 1+i \end{array} \right]. $$ Therefore, $$ M^{-1} = \frac{1}{4i} \left[ \begin{array}{cc} 1+i & i-1 \\ i-1 & 1+i \end{array} \right]. $$ It corresponds to a Mobius transform, $$ f(z) = \frac{(1+i)z + (i-1)}{(i-1)z + (1+i)}. $$
You can check that it is what you want at the beginning.