I want to find the Möbius transformation $f$ such that
$f(0)=0,f(1)=1,f(\dfrac12+\dfrac{\sqrt{3}}{2}i)=(0.4387+0.759775 \ i)$
Is this possible? I tried both cross ratios and systems of equation approaches but I can't find any solution. Thanks for helps.
A general Möbius transformation has the form $T(z) = (az+b)/(cz+d)$, and we can take $ad-bc=1$. If $T(0)=0$, we have $$0=\frac{b}{d},$$ so $b=0$ (and $d \neq 0$, or $T(0) = \infty)$. Hence $a \neq 0$, so divide by it and look for $T(z) = 1/(Cz+D) $.
$T(1)=1$, so $$ 1 = \frac{1}{C+D}. $$ or $$ C+D=1. \tag{1}$$ Finally, if you want to send $\alpha$ to $\beta$, you need $$ \beta = \frac{\alpha}{C\alpha+D}, $$ or $$ C\alpha+D=\frac{\alpha}{\beta} \tag{2} $$ You can then solve (1) and (2) simultaneously, which will work since $\alpha \neq 1$. You'll find $$ T(z) = \frac{(1-\alpha)z}{(1-\beta/\alpha)z+(\beta/\alpha-\alpha)}. $$