I'm working my way through Martin Schechter's 'Principles of Functional Analysis' (2nd ed.) and am trying to understand his proof of the following theorem, given on page 136:
"Let $T:X\to X$ be any bounded linear operator on a complex Banach space $X$ and let $C$ be any closed curve containing $\sigma(A)$ on its interior. Then $$T^{n}=\frac{1}{2\pi i}\oint_{C}z^{n}\left(zI-T\right)^{-1}dz."$$
The proof uses the following Lemma, which I think I understand:
"Let $T:X\to X$ be a bounded linear operator on a complex Banach space $X$ and $z\in\mathbb{C}$ be s.t. $\left|z\right|>\limsup\left\Vert T^{n}\right\Vert ^{\frac{1}{n}}$. Then $$\left(zI-T\right)^{-1}=\sum_{n=1}^{\infty}z^{-n}T^{n-1},$$ where the RHS converges in the norm on the set of bounded operators."
The proof of the theorem in hand goes like this:
Let $C$ be any circle of radius greater than $\left\Vert T\right\Vert$ centred at the origin. Then, using the Lemma,
$\oint_{C}z^{n}\left(zI-T\right)^{-1}dz=\oint_{C}z^{n}\sum_{k=1}^{\infty}z^{-k}T^{k-1}dz=\sum_{k=1}^{\infty}T^{k-1}\oint_{C}z^{n-k}dz,$
all of which I follow, but then Schechter goes onto to assert that $\sum_{k=1}^{\infty}T^{k-1}\oint_{C}z^{n-k}dz = 2 \pi i T^n$, which I don't understand. Can anyone help me out?
(For anyone interested, to conclude, Schechter argues that since line integrals are path independent, and $\left(zI - T \right)^{-1}$ is analytic on $\rho (T)$, the theorem holds.)
By Cauchy's integral formula, $$\oint_{C}z^{n-k}dz=\left\{\begin{array}{cc}2\pi i & k=n+1\\0&\text{otherwise}\end{array}\right..$$ Alternatively, since $C$ is a circle centered at $0$, you can let $z=re^{i\theta}$ to evaluate the integral above, where $r$ is the radius of $C$.