Solution
= Solution
Divide the complex <polynomial> by its known factor $z-(2+i)$:
$$
p(z)=(z-2-i)\big[z^2-(2+i)z+(1+i)\big].
$$
The quadratic factors further as $(z-1)(z-1-i)$, because its two proposed roots sum to $2+i$ and multiply to $1+i$. Therefore the remaining solutions are
$$
\boxed{z=1,\qquad z=1+i.}
$$
All three roots are distinct. Complex coefficients do not require nonreal roots to occur in conjugate pairs.