Construct a Mobius transformation $f$ with the specified effect: $f$ maps $K(0,1)$ to itself and $K(1/4,1/4)$ to $K(0,r)$ for some $r<1$.
My work: Since
$i \rightarrow 1 \leftarrow i$
$ 1/4 \rightarrow 0 \leftarrow 0$
$ 0 \rightarrow \infty \leftarrow -r$
Define linear fractional transformation as
$T_1(z)= ( \frac{i}{i-\frac{1}{4}} )\frac{z-\frac{1}{4}}{z}$,
$T_2(z)= ( \frac{z}{z+r} )\frac{i+r}{i}$
And so, we get $T_2^{-1}(z)=\frac{i rz}{i+r-iz}$
Therefore, $f(z)=T_2^{-1}\circ T_1(z)= \frac{i r ( \frac{i}{i-\frac{1}{4}} )\frac{z-\frac{1}{4}}{z}}{i+r-i( \frac{i}{i-\frac{1}{4}} )\frac{z-\frac{1}{4}}{z}}$
Did I miss anything? I really appreciate your kind help. Thank you!
Assuming that “$K(a,b)$” is the circle centered at $a$, with radius $b$, I think that your method was probably all right, but that you made things more difficult for yourself by supposing that $i$ was a fixed point on the unit circle.
It makes much more sense to treat your fixed points as $\pm1$, and in that case, since you want $0\mapsto-r$ and $1/2\mapsto+r$, the whole problem becomes a question on fractional-linear transformations with real coefficients.
After too much work, I found that the Möbius transformation leaving $\pm1$ fixed and sending $0$ to an unspecified $-r$ is $$ z\mapsto\frac{z-r}{-rz+1}\,. $$ I used exactly the technique that you did for your choice of specified points, so I won’t go through the computation here.
And you want to adjust your $r$ so that $1/2$, the other real point on $K(\frac14,\frac14)$, goes to $+r$. You get \begin{align} \frac12&\mapsto\frac{\frac12-r}{-\frac r2+1}=\frac{1-2r}{-r+2}=r\\ r(2-r)&=1-2r\\ 0&=1-4r+r^2\\ r&=2\pm\sqrt3\,, \end{align} where of course it’s the minus sign that must be chosen. And surenough, if we call $2-\sqrt3=u$, we have, if $f(z)=(z-u)\big/(-uz+1)$, then $f(-1)=-1$, $f(0)=-u$, $f(1/2)=u$, $f(1)=1$.