Solution
= Solution
The fixed-point equation is
$$
\frac{az+b}{z+1}=z
\quad\Longleftrightarrow\quad
z^2+(1-a)z-b=0.
$$
A unique solution at $z=i$ means that this <polynomial> is
$$
(z-i)^2=z^2-2iz-1.
$$
Comparing coefficients gives
$$
\boxed{a=1+2i,\qquad b=1}.
$$
This is the repeated-root case of <prescribed fixed points of a Möbius transformation>.