I'm following Martin Schechter's 'Principles of Functional Analysis' (Second Edition, 2002) and am interested in the spectral theory chapter (chapter six).
In particular, I wish to make us of Theorem 6.12 on page 136, which states:
"Let $C$ be any closed curve containing $\sigma (A)$ in its interior. Then $$A^n = \frac{1}{2 \pi i} \oint_C z^n (z - A )^{-1} dz. $$"
Here, I believe $A$ is a bounded operator over $X$, a complex Banach space; the line integral is taken in the right direction; and $\sigma(A)$ is the spectrum of $A$, that is all the points not in the resolvent, i.e. all those points $\lambda$ for which there does not exist a bounded inverse for the operator $A - \lambda I$.
Now, I want to check this result with a simple example. Namely, I know that for any diagonal matrix, $ A = \left[ \begin{matrix} a_1 & 0 \\ 0 & a_2 \end{matrix}\right] $, we have $A^n = \left[ \begin{matrix} a_1^n & 0 \\ 0 & a_2^n \end{matrix} \right]$. But how to prove this using the aforementioned theorem?
Well, $A - \lambda I = \left[ \begin{matrix} a_1 - \lambda & 0 \\ 0 & a_2 - \lambda \end{matrix}\right]$ does not have a bounded inverse if $ \det (A - \lambda I ) = \left(a_1 - \lambda\right)\left(a_2 - \lambda \right) = 0$. Thus, $\sigma(A) = \{a_1, a_2 \} $. So, we set $C$ to be any closed curve containing the points $a_1$ and $a_2$ in the complex plane in its interior; for instance, I suppose we could set $C$ to be the circle centred at the origin with radius $r := \max \{ \left|a_1 \right|, \left|a_2\right| \}$.
Now
$$ \left(z - A \right)^{-1} = \left[ \begin{matrix} z - a_1 & 0 \\ 0 & z - a_2 \end{matrix}\right]^{-1} = \left[ \begin{matrix} \frac{1}{z - a_1} & 0 \\ 0 & \frac{1}{z - a_2} \end{matrix}\right],$$
so
$$ z^n \left(z - A \right)^{-1} = \left[ \begin{matrix} \frac{z^n}{z - a_1} & 0 \\ 0 & \frac{z^n}{z - a_2} \end{matrix}\right].$$
However, I'm not sure where to take things from here to get the desired result. Can anyone help me?