In Silverman's complex analysis textbook, he described how to find Mobius transformation that maps $|z+i|=1$ to $|w|=4$.
Let $T:\Bbb C_\infty\to\Bbb C_\infty$ be a Mobius transformation that we require. Then $Tz = \frac{az+b}{cz+d}$ for some $a,b,c,d\in\Bbb C$ such that $ad-bc \neq 0$. WLOG, let $T(-i) = \infty$. Then we can reduce the form to $Tz =\frac{az+b}{z+i}$. Since two poins $0,-2i$ in $|z+i| = 1$ should maps to points in $|w|=4$ and $T(0) = -ib$ and $T(-2i) = 2a+ib$, $|b| = 4$ and $\sqrt{4a^2+b^2} = 4$. Hence, we may choose $b = 4$ and $a = 0$. This gives $T(z) = \frac{4}{z+i}$. And then the book conclude that this is the desired map. I understand that $T$ maps circle to circle but circle that passing $4i$ and $-4i$ are infinitely many. How did he ensure this maps $|z+i| = 1$ to $|w| = 4$?
Note that if $|z+i|=1$, then $|T(z)| = \left|\frac 4{z+i}\right| = 4$.