I am reading Axler's Linear Algebra Done Right and he proved every invertible linear complex operator has a square root. Following the prove one can show that they have any n-th root, hence I wonder if the stronger algebraically closed condition fulfills too. I have absolutely no idea how to tackle such problem.
In other words, let $T$ be an invertible operator in $\mathcal L(V)$, $P$ be a complex polynomial, is there an invertible operator $S$ such that $P(S)=T$.
