How to apply Cauchy's formula in proof 10.13 of Rudin's FA Let $A$ be a Banach algebra and $x\in A$.
In part of prood 10.13 of Rudin's Functional Analysis (page 254), where he is trying to prove that $$\rho(x) = \lim_{n\to\infty}||x^n||^{\frac{1}{n}} = \inf\left(\{||x^n||^{\frac{1}{n}}\}|n\geq1\right)$$ (where $\rho(x)$ is the spectral radius of $x$) he wants to apply the Cauchy theorem (3.31) to the map $f:(\mathbb{C}\backslash\sigma(x))\to G(A)$ given by $$\lambda\mapsto (\lambda e-x)^{-1}$$ (where $G(A)$ is the group of invertible elements of $A$) in order to obtain that for all $r>\rho(x)$  the following formula is fulfilled $$ x^n = \frac{1}{2 \pi i}\int_{\partial B_r(0)} \lambda^n f(\lambda) d\lambda $$
However, I don't understand how the Cauchy formula (theorem 3.31) can be applied in this case, because it says that $$ f(z) = \frac{1}{2\pi i}\int_{\partial B_r(0)} \frac{f(\lambda)}{\lambda-z}d\lambda $$ which would allow you to conclude that $$ (z e - x)^{-1} = \frac{1}{2\pi i}\int_{\partial B_r(0)} \frac{f(\lambda)}{\lambda-z}d\lambda $$
It feels to me like he is actually applying the Cauchy formula on another map, $h:A\to A$, given by $x\stackrel{h}{\mapsto} x^n$. But in order to do this, you must have the following, modified Cauchy formula: $$ h(x) = \frac{1}{2\pi i} \int_{\partial B_r(0)} h(\lambda)(\lambda e -x)^{-1}d\lambda $$
I realize such a formula probably does exist, but it is not 3.31. What am I missing?
Also, how did he get (in proof 10.13) from (9) to (10)? Does he need to assume that $r M(r) > 1$?
 A: The part of theorem 3.31 that is used here is the equality $(3)$; for closed paths $\Gamma_1,\,\Gamma_2$ that are homologous in $\Omega$ (that is, they have the same winding number around all $w \in \mathbb{C}\setminus \Omega$), and any (weakly) holomorphic function $f$ on $\Omega$ we have
$$\int_{\Gamma_1}f(\zeta)\,d\zeta = \int_{\Gamma_2} f(\zeta)\,d\zeta.\tag{3}$$
In our situation, we have $\Omega = \mathbb{C}\setminus \sigma(x)$ and $f(\lambda) = \lambda^n (\lambda e - x)^{-1}$. For $\lvert\lambda\rvert > \lVert x\rVert$, we have the expansion
$$(\lambda e - x)^{-1} = \lambda^{-1}(e - \lambda^{-1}x)^{-1} = \lambda^{-1}\sum_{n = 0}^\infty \frac{x^n}{\lambda^n} = \sum_{n = 0}^\infty \frac{x^n}{\lambda^{n+1}},$$
and the convergence of the series is uniform on all $\mathbb{C}\setminus D_R(0)$ with $R > \lVert x\rVert$, so for $R > \lVert x\rVert$ we have
$$\int_{\lvert \lambda\rvert = R} f(\lambda)\,d\lambda = \sum_{k = 0}^\infty \int_{\lvert\lambda\rvert = R} \lambda^{n- k - 1}\,d\lambda\cdot x^k = 2\pi i x^n.$$
We fix an $R > \lVert x\rVert$. For any $r > \sigma(x)$, the two closed curves $\Gamma_r$ and $\Gamma_R$, where $\Gamma_\rho(t) = \rho\cdot e^{it}$, are homologous in $\Omega$ (they are even homotopic as loops), so by $(3)$, we have
$$\frac{1}{2\pi i} \int_{\Gamma_r} f(\lambda)\,d\lambda = \frac{1}{2\pi i} \int_{\Gamma_R} f(\lambda)\,d\lambda = x^n.$$

From the inequality $(9)$,
$$\lVert x^n\rVert \leqslant r^{n+1} M(r),\tag{9}$$
by taking $n^{\text{th}}$ roots - all involved quantities are non-negative - we obtain
$$\lVert x^n\rVert^{\frac{1}{n}} \leqslant r\cdot \bigl(rM(r)\bigr)^{\frac{1}{n}}.\tag{$9'$}$$
Now $0 < rM(r) < +\infty$, and therefore
$$\lim_{n\to\infty} \bigl(rM(r)\bigr)^{\frac{1}{n}} = 1.\tag{$\ast$}$$
So, given $\varepsilon > 0$, from $(\ast)$ and $(9')$ we obtain
$$\lVert x^n\rVert^{\frac{1}{n}} \leqslant r\cdot (1+\varepsilon)$$
for all large enough $n$. This implies
$$\limsup_{n\to\infty} \lVert x^n\rVert \leqslant r\cdot (1+\varepsilon),$$
and since $\varepsilon > 0$ was arbitrary, we finally obtain
$$\limsup_{n\to\infty} \lVert x^n\rVert^{\frac{1}{n}} \leqslant r.\tag{10}$$
