For $\lambda\in\mathbb{R}$, I want to find the expression of $f(\lambda)$: $$f(\lambda)=\max_{E\in\mathbb{C}}arccosh{\frac{|E^2+i\lambda E|+|E^2+i\lambda E-4|}{4}}-arccosh{\frac{|E^2-i\lambda E|+|E^2-i\lambda E-4|}{4}}$$ that is $$A={\sqrt{a^2+b^2} \sqrt{a^2+(b+\lambda )^2}+\sqrt{\left(a^2-b^2-b \lambda -4\right)^2+(2 a b+a \lambda )^2}+\sqrt{\left(\sqrt{a^2+b^2} \sqrt{a^2+(b+\lambda )^2}+\sqrt{\left(a^2-b^2-b \lambda -4\right)^2+(2 a b+a \lambda )^2}\right)^2-16}}$$ $$B={\sqrt{a^2+b^2} \sqrt{a^2+(b-\lambda )^2}+\sqrt{\left(a^2-b^2+b \lambda -4\right)^2+(2 a b-a \lambda )^2}+\sqrt{\left(\sqrt{a^2+b^2} \sqrt{a^2+(b-\lambda )^2}+\sqrt{\left(a^2-b^2+b \lambda -4\right)^2+(2 a b-a \lambda )^2}\right)^2-16}}$$ $$f(\lambda)=\max_{a\in\mathbb{R},b\in\mathbb{R}}\log\frac{A}{B} $$
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$\begingroup$ I took the liberty to remove the "non-standard analysis" tag, since this has a particular technical sense that (I strongly suspect) is not at all related to what you're expecting in a response. (It refers to model-theoretic or other versions of real numbers which allow genuine infinitesimals, and develop understanding of how to use them correctly... It's not just "unusual analysis"...) $\endgroup$– paul garrettCommented Nov 5, 2021 at 23:04
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As $g(t)=\log{t}$ is a strictly increasing function, what you really needs is just the maximum of $\frac{A}{B}.$
Also, after some experiments, it seems that, for pure imaginary numbers only, the maximum occurs when the imaginary part is equal to $\lambda$. That is (in your notation), when $a=0$ the maximum is obtained with $b=\lambda$. So "divide and conquer", by restricting at one curve at a time, may be a efficient approach to the full solution.
