Solution
= Solution
For roots $3i$ and $1+i$, the fixed-point <polynomial> must be
$$
(z-3i)(z-1-i)
=z^2-(1+4i)z+(3-3i).
$$
Comparing this with $z^2+(1-a)z-b$ gives
$$
\boxed{a=2+4i,\qquad b=-3+3i}.
$$
= Solution
For roots $3i$ and $1+i$, the fixed-point <polynomial> must be
$$
(z-3i)(z-1-i)
=z^2-(1+4i)z+(3-3i).
$$
Comparing this with $z^2+(1-a)z-b$ gives
$$
\boxed{a=2+4i,\qquad b=-3+3i}.
$$