For $z$ near infinity (sufficiently large $z$), I want to verify the following claims :

  1. If $f$ is univalent (an analytic injective function) such that $f(z)=z+O(1/z)$, then $f^{-1}(z)=z+O(1/z)$.

  2. $\log|z+O(1)|=\log|z|+o(1)$ (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!

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  • $\begingroup$ Thanks, I have a few things to clarify. First, |w|>|z|-C/|z|=(|z|^2-C)/|z|>C/|z| so how can you end up with |z|/2 on the second inequality. Well, I think you are using the reverse triangle inequality to get |w-z|<2C/|w| but I cannot see how. Finally, for #2 I guess you use log|1+x|<2|x|, |x|<1/2 (I trust this from the graph). Does it matter to check that |o(1)|<1/2? Well, from the definition we cannot tell, so I'm curious how do we properly apply the inequality. $\endgroup$ – Nothingone May 20 at 4:03
  • $\begingroup$ if $|z|>R, R^2>2C$ then $|z|- C/|z| \ge |z|/2$ by a simple check (pass $|z|/2$ to the left, $-C/|z|$ to the right, simplify etc); for the last $o(1)$ means a small quantity that goes to zero in absolute value) so it is automatically less than any fixed $\delta >0$ so in particular $1/2$; the logarithmic inequality follows by the Taylor series immediately by majorizing that with a geometric series- if you are not sure about these notations, try and redo this with limits and inequalities; it's messier and longer but it's worth doing it at least once to undertsand the usage of $O,o$; $\endgroup$ – Conrad May 20 at 11:31

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