For $\lambda\in\mathbb{R}$, I want to find the expression of $f(\lambda)$: $$f(\lambda)=\max_{E\in\mathbb{C}}arccosh{\frac{E^2+i\lambda E+E^2+i\lambda E4}{4}}arccosh{\frac{E^2i\lambda E+E^2i\lambda E4}{4}}$$ that is $$A={\sqrt{a^2+b^2} \sqrt{a^2+(b+\lambda )^2}+\sqrt{\left(a^2b^2b \lambda 4\right)^2+(2 a b+a \lambda )^2}+\sqrt{\left(\sqrt{a^2+b^2} \sqrt{a^2+(b+\lambda )^2}+\sqrt{\left(a^2b^2b \lambda 4\right)^2+(2 a b+a \lambda )^2}\right)^216}}$$ $$B={\sqrt{a^2+b^2} \sqrt{a^2+(b\lambda )^2}+\sqrt{\left(a^2b^2+b \lambda 4\right)^2+(2 a ba \lambda )^2}+\sqrt{\left(\sqrt{a^2+b^2} \sqrt{a^2+(b\lambda )^2}+\sqrt{\left(a^2b^2+b \lambda 4\right)^2+(2 a ba \lambda )^2}\right)^216}}$$ $$f(\lambda)=\max_{a\in\mathbb{R},b\in\mathbb{R}}\log\frac{A}{B} $$

$\begingroup$ I took the liberty to remove the "nonstandard 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 modeltheoretic 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 garrettNov 5, 2021 at 23:04
1 Answer
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.