Find the transformation that maps real axis to itself and imaginary axis to the circle $|w-\frac{1}{2}|=\frac{1}{2}$
What I did: $$z_{1}=0,z_{2}=i,z_{3}=\infty ,w_{1}=0,w_{2}=\frac{1}{2}(1+i),w_{3}=1$$ $$f_{1}(z)=\frac{z-z_{1}}{z-z_{3}}\frac{z_{2}-z_{3}}{z_{2}-z_{1}}=\frac{z}{i}$$ $$f_{2}(w)=\frac{w-w_{1}}{w-w_{3}}\frac{w_{2}-w_{3}}{w_{2}-w_{1}}=\frac{wi}{w-1}$$ $$f^{-1}_{2}(w)=\frac{z'}{z'-i}$$ $$T(z)=f^{-1}_{2}(f_{1}(z))=\frac{\frac{z}{i}}{\frac{z}{i}-i}=\frac{z}{z+1}$$ But the answer should be $T(z)=\frac{1}{z+1}$, does anyone could help me to find the mistake I made? Thanks very much!
Begin by considering any number in the imaginary axis, say $ai$ for $a\in\mathbb{R}$, and plug it into your function: $$ \frac{ai}{1+ai} = \frac{ai(1-ai)}{1+a^2} = \frac{a(a+i)}{1+a^2}. $$ Then, let's see its distance from $1/2$: $$ \left|\frac{a(a+i)}{1+a^2}- \frac{1}{2} \right|^2= \left|\frac{2a(a+i)-1-a^2}{2(1+a^2)}\right|^2 = \left|\frac{a^2+2ai-1}{2(1+a^2)}\right|^2= \frac{((a^2-1)^2 +4a^2)}{4(1+a^2)^2}=\frac{1}{4}. $$ Which means your transformation actually sends the imaginary axis to the circle (and it obviously maps the real axis to itself), so there's nothing wrong here. The thing is the map you're asked for is not unique.