# Proof that $A^n = \frac{1}{2\pi i} \oint _C z^n \left(z - A \right)^{-1} dz$

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 $C$.
• Thank you! Should that be $n=k-1$? – Harry Williams Apr 28 '13 at 18:05
• I've tried to get out the result using your suggestion of letting $z=re^{i\theta}$, but I'm struggling. In this case, I think we get $\oint_C z^{n-k} dz = \int_0^{2 \pi} r^{n-k} e^{i (n-k) \theta} d \theta = r^{n-k} \left( \int_0^{2 \pi} \cos \left( (n-k) \theta \right) d \theta + i \int_0^{2 \pi} \sin \left( (n-k) \theta \right) d \theta \right)$ but I seem to get that this is always zero. Where am I going wrong? – Harry Williams Apr 28 '13 at 18:56
• @HarryWilliams: Note that $dz=ire^{i\theta}d\theta$. – Hu Zhengtang Apr 28 '13 at 19:02