Find a Möbius tranformation f such that:
f(0)=1
f'(0)=2
f''(0)=3
I know the solution is: f(z)= (5z+4)/(4-3z) but I don't know how to get to it. Thanks for you help!
You know, I hope, that a Moebius transformation is of the form $\displaystyle f(z)=\frac{z+ b}{cz+ d}$ (you appear to have it in the form $\displaystyle \frac{a'z+ b'}{c'z+ d'}$ but my coefficients are the result of dividing both numerator and denominator by a').
From that determine f'(z) and f''(z). Setting f(z)= 1, f'(z)= 2, and f''(z)= 3 gives three equations to solve for b, c, and d.