Is the matrix ring $\mathrm{Mat}_n(\mathbb{C})$ "algebraically closed"?

The matrix $\left( \begin{array}{cc} 0 & 1 \\\\ 0 & 0 \end{array} \right)$ has no square root.

Polynomials make sense for continuous complex functions on a space. If that space is $\mathbb R$, then polynomial equations with complex coefficients are solvable. If that space is $\mathbb C$ or $S^1$ then $g^2 = f$ may not be solvable.

I'd like to add that a nice theory of roots of polynomials over noncommutative rings was developed by I. Gelfand, V. Retakh, and R. Wilson, see the paper arXiv:math/0208146 and references therein (in particular, the earlier paper by Gelfand and Retakh on the noncommutatoive Vieta theorem).