I feel this is not hard, but no way to prove it
$|\sqrt{z^2 -4}-z|\le 2$
Any body can help? Thanks!
The total statement should be one of the branchs of square root should satisfy this inequality.
2026-04-23 01:07:35.1776906455
A simple complex inequality
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Since $$(z+\sqrt{z^2-4})(z-\sqrt{z^2-4}) = 4$$ one of the factors has absolute value at most $2$. That is, inequality holds for a choice of the branch of square root. You cannot make this choice consistently on the entire plane, but you can make it on $\mathbb C\setminus [-2,2]$.
This is related to inverting the Joukowski map $z+1/z$.