On page 33 of http://math.aalto.fi/opetus/harmanal/pruju/calg04.pdf it is asked in what kind of Banach algebras is 0 the only topological zero divisor.
What do they mean by kind of Banach algebras. Using part (b) of the exercise it follows that the Banach algebras may not have invertible elements. Is this a kind of Banach Algebras?
Writing $G(A)$ to be the open subset of invertible elements, any $x \in \partial G(A)$ (the boundary) is a topological zero divisor. To see this, just observe that $\frac{xx_n^{-1}}{\|x_n^{-1}\|}\to 0$ whenever $x_n\in G(A)$ converges to $x$.
If $A$ has no nonzero topological zero divisors then, by the above, $\{0\}\cup G(A)$ is closed and so the set of nonzero non invertible elements, call it $S(A)$, is open. Further, if $x \in S(A)$, then the continuous function $f(\lambda)= (1-\lambda)x + \lambda e$ on $[0,1]$ has image in $S(A) \cup G(A)$, the union of two disjoint open sets. Contradiction. So the assumption that $S(A) \neq \emptyset$ was wrong.
Hence $A = \mathbb{C}$ by the Gelfand-Mazur theorem.