# Function (Taylor series) of Jordan canonical form about arbitrary point

On the Wikipedia page for Jordan matrices, under the section on functions $$f$$ of matrices $$A = PJP^{-1}$$ (that being the Jordan Canonical Form), the following can be found (I've paraphrased it a tiny bit to make it concern my point):

Let $$f(z)=\sum _{{n=0}}^{{\infty }}a_{n}(z-z_{0})^{n}$$ be the power series expansion of $$f$$ around $$z_{0}\in\mathbb{C}$$, which will be hereinafter supposed to be $$0$$ for simplicity's sake. The matrix $$f(A)$$ is then defined via the following formal power series: $$f(A)=\sum _{n=0}^{\infty }a_{n}A^{n}$$.

Obviously, scalar functions of matrices can then be computed via $$f(A) = P\Big(\sum^\infty_{n=0}a_nJ^n\Big)P^{-1}$$, thanks to

$$A^n = PJP^{-1}PJP^{-1} ... PJP^{-1}PJP^{-1} = PJ^nP^{-1}.$$

However, that's assuming $$z_0=0$$. You might want to use a Taylor series (since, in well-behaved cases, such a series exists at any point on the domain of $$f$$) about an arbitrary point $$z_0$$. The Wikipedia page doesn't cover that case, however. That'd mean the summation would be summing matrix powers of the form:

$$(A-z_0)^n = (PJP^{-1}-z_0)(PJP^{-1}-z_0) ... (PJP^{-1}-z_0)(PJP^{-1}-z_0) = \;?$$

... where, I guess, $$z_0\equiv z_0I$$. My question is two-fold:

1. Do those powers amount to anything as nicely separable as when $$z_0=0$$?
2. Say I want to compute the value of $$f(A)$$ with $$f(x) = (1-x)^{-1}$$, where $$A$$ has only one eigenvalue, that is confined to $$]-1,1[$$. Would it make sense at all to use the Taylor series about that eigenvalue, instead of the Maclaurin series? Furthermore, say $$A$$ had many different eigenvalues, possibly on intervals certain Taylor series of $$f$$ don't converge on. Is there any guideline as to which point we'd choose the series about, in that case?