Let $A$ be a Banach algebra and $a\in A$. $\|z(z-a)^{-1}\|=1+o(\frac{1}{z})$ as $z\rightarrow +\infty$ iff $\|(z-a)^{-1}\|^{-1}=z+o(1)$ as $z\rightarrow +\infty$. i.e., $lim_{z\rightarrow +\infty}\|(z-a)^{-1}\|^{-1}-z=0$ iff $lim_{z\rightarrow\infty}\frac{\|z(z-a)^{-1}\|-1}{\frac{1}{z}}=0$.
One way follows easily.
Suppose $lim_{z\rightarrow\infty}\frac{\|z(z-a)^{-1}\|-1}{\frac{1}{z}}=0$.
Then $lim_{z\rightarrow\infty}\|z(z-a)^{-1}\|-1=0$ $\Rightarrow lim_{z\rightarrow +\infty}\frac{1}{\|(z-a)^{-1}\|}-z=0$ $\Rightarrow \|(z-a)^{-1}\|^{-1}=z+o(1)$ as $z\rightarrow +\infty$.
How does the converse follow? I would be grateful for a hint. Thanks in advance.
Edit: Here is what I've got- do let me know if this seems correct!
Suppose $\|(z-a)^{-1}\|^{-1}=z+o(1)$ as $z\rightarrow +\infty$. i.e., $lim_{z\rightarrow +\infty}\frac{1}{\|(z-a)^{-1}\|}-z=0 \Rightarrow lim_{z\rightarrow +\infty}\frac{1-\|z(z-a)^{-1}\|}{\|(z-a)^{-1}\|}=0$ $\Rightarrow lim_{z\rightarrow +\infty}\frac{\|z(z-a)^{-1}\|-1}{\|(z-a)^{-1}\|}=0$.
Now $lim_{z\rightarrow +\infty}\frac{\|z(z-a)^{-1}\|-1}{\frac{1}{z}}=$ $lim_{z\rightarrow +\infty}\frac{\|z(z-a)^{-1}\|-1}{\|(z-a)^{-1}\|}\frac{\|(z-a)^{-1}\|}{\frac{1}{z}}$.
Consider $lim_{z\rightarrow +\infty}\frac{1}{\|(z-a)^{-1}\|}-z=0$, by hypothesis.
$\Rightarrow lim_{z\rightarrow +\infty}\frac{1}{z\|(z-a)^{-1}\|}=1$
Hence $lim_{z\rightarrow +\infty}\frac{\|z(z-a)^{-1}\|-1}{\frac{1}{z}}=0\cdot1=0$, as required.
Hence $\|z(z-a)^{-1}\|=1+o(\frac{1}{z})$ as $z\rightarrow +\infty$.