= Solution
{c}
The prescribed <pole> lets us choose $c=4$ and $d=-5-3i$. Write $f(z)=(Az+B)/(4z-5-3i)$. Fixing $1/2$ and $2$ gives
$$
A+2B=-3-3i,\qquad 2A+B=6-6i,
$$
so $A=5-3i$ and $B=-4$. Thus
$$
\boxed{f(z)=\frac{(5-3i)z-4}{4z-5-3i}.}
$$
Its <determinant> is $-18\ne0$, so it is a genuine <Möbius transformation>. Direct substitution gives $f(1/2)=1/2$, $f(2)=2$, and a pole at $(5+3i)/4$; the numerator at that pole is $9/2\ne0$.
To identify the <image> of the <unit disc>, put
$$
\alpha=\frac4{5-3i}=\frac{10+6i}{17},\qquad
\lambda=-\frac{5-3i}{5+3i}.
$$
Then $|\alpha|^2=8/17<1$, $|\lambda|=1$, and
$$
f(z)=\lambda\frac{z-\alpha}{1-\overline\alpha z}.
$$
In particular, its pole is outside the closed <unit disc>. The identity
$$
1-|f(z)|^2=\frac{(1-|\alpha|^2)(1-|z|^2)}{|1-\overline\alpha z|^2}
$$
shows $f(D)\subseteq D$. For any $|w|<1$, solve $w=\lambda(z-\alpha)/(1-\overline\alpha z)$ to get
$$
z=\frac{w/\lambda+\alpha}{1+\overline\alpha w/\lambda}.
$$
Its denominator is nonzero, and the same <modulus> identity for this inverse gives $|z|<1$. Thus $f(D)=D$, and global <injectivity> of a <Möbius transformation> proves the required one-to-one correspondence.
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