 Solution

The Artin–Wedderburn theorem says that a finite-dimensional semisimple algebra over $\mathbb{C}$ is a finite product ${\prod }_j{M}_{d_j}(\mathbb{C})$: the division algebras in the general theorem reduce to $\mathbb{C}$ because the base field is algebraically closed. Each factor has center of an associative algebra $\mathbb{C}{I}_{d_j}$. Consequently a central element corresponds to $({\lambda }_j{I}_{d_j}{)}_j$. If it is nilpotent, then ${\lambda }_j^n=0$ for every $j$, whence every ${\lambda }_j=0$. Therefore $\boxed{z=0}$. The centrality assumption matters: a full matrix algebra has nonzero nilpotent elements.