There's a statement in Rudin's functional analysis stating the following in the context of Symbolic Calculus in the Banach Algebra chapter:
If $x$ is an element of a Banach Algebra $A$ and if $f(\lambda) = \alpha_0 + \ldots \alpha_n \lambda^n$ is a polynomial with complex coefficients $\alpha_i$, there can be no doubt about the meaning of the symbol $f(x)$; it obviously denotes the element of $A$ defined by $$ f(x) = \alpha_0 e + \ldots \alpha_n x^n $$ The question aries whether f(x) can be defined in a meaningfull way for other functions $f$. We have already encountered some examples of this. For instance during the proof ot Theorem 10.9 we came very close to defining the exponential function in $A$. In fact if $f(\lambda) = \sum \alpha_k \lambda^k$ is any entire function in $\mathbb{C}$, it is natural to define $f(x) \in A$ by $f(x) = \sum \alpha_k x^k$; this series always converges. Another example is given by the meromorphic functions $$ f(\lambda) = \frac{1}{\alpha - \lambda}. $$ In this case, the natural definition of $f(x)$ is $$ f(x) = (\alpha e - x)^{-1} $$ which makes sense for all $x$ whose spectrum does not contain $\alpha$. One is thus led to the conjecture that $f(x)$ should be definable, within $A$, whenever $f$ is holomorphic in an open set that contains $\sigma(x)$. This turns out to be correct and can be accomplished by a version of the Cauchy formula that converts complex functions defined in open subsets of $\mathbb{C}$ to A-valued ones defined in certain open subsets of $A$. (Just as in classical analysis, the Cauchy formula is a much more adaptable tool than the power series representation).
Why is it true that the Cauchy integral formula is better than the power series representation? Is it because we can represent functions with singularities as well while the Taylor series for example cannot? I know what Rudin is trying to achieve, defining functions of elements of Banach Algebra given complex functions but I just don't quite get the subtlety of that statement.