For $z$ near infinity (sufficiently large $z$), I want to verify the following claims :
If $f$ is univalent (an analytic injective function) such that $f(z)=z+O(1/z)$, then $f^{-1}(z)=z+O(1/z)$.
$\log|z+O(1)|=\log|z|+o(1)$ (How does this identity involve both Big-Oh and little-oh?)
Well, I try to using power series expansion for $\log(1+x)$ but it does not seem work.
For 1, let $|z|\ge R>0$ with $R$ large tbd and $w=f(z)$
$|w-z| \le C/|z|$ means $|w| \le |z|+ C/|z| \le 2|z|$ as well as $|w| \ge |z|- C/|z| \ge |z|/2$ if we choose $R^2 \ge 2C$ hence we have $|w-z| \le 2C/|w|, |z| >R$ as above and $O(1/w)=O(1/z)$
Since $z=f^{-1}(w)$ we get $|w-f^{-1}(w)|=O(1/w)$ or $f^{-1}(w)=w+O(1/w)$ Changing variables back to $z$ we are done!
For 2 we have $\log|z+O(1)|=\log |z|+\log |(1+O(1)/z)|$
But $O(1)/z=o(1)$ for $|z|$ large and $\log |1+x| \le 2|x|, |x| \le 1/2$, so $\log |(1+O(1)/z)|=\log |(1+o(1)|=o(1)$ and putting all together we get
$\log|z+O(1)|=\log |z|+o(1)$ so we are done!