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The question might seem basic and perhaps I am overlooking something. I'm looking at the development of the extremum of $$ f(x)=\left|\frac{ A \mathbf{x} \sin\left[B\sqrt{x^2 -a^2+2 i \cos[\Phi ] a b+b^2}\right] }{\sqrt{x^2 -a^2+2 i \cos[\Phi ] a b+b^2}}+\text{ }i \cos\left[ B \sqrt{x^2 -a^2+2 i \cos[\Phi ] a b+b^2}\right]\right|^2 $$ ($A,B,a,b>0$) that lies at $x=0$ when $\Phi=\pi/2$ and shifts to the left for smaller $\Phi$ and to the right for larger ones or vice versa depending on the parameters.

$a$ may be larger than $b$ to keep this effect, so the root may be completely imaginary and completly real depending on $x$ when $\cos[\Phi]=0$. The point may lie in the mixed terms occuring when $\cos[\Phi] \neq 0$, but I can't quite get a hold on it.

However, the term is too complicated for simply setting the derivative to zero and solving for $x$.

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You seem to say that $x=0$ is an extremum of $f$ when $\cos\Phi=0$. You could explain why. – Did Oct 28 '11 at 11:46
Because the physical experiment lieing behind this formula says so, at least for a wide range of parameters. :) But I also think it should come out easier as the root is either imaginary or real then. Unfortunately I pressed enter for this comment but I will check and post another one. – mcandril Oct 28 '11 at 11:59
It might be an excellent idea to mention what this "physical experiment" you speak of is. – J. M. is back. Oct 28 '11 at 12:00
!!!CORRECTED ERROR (bold) IN FORMULA, SORRY!!! Well, as the derivation of the formula is not published until now, I wouldn't want to go too much into detail :) I don't think it will give more inside than just plotting the formula. Take {A -> 1, B -> 1, a -> .002, b -> .001} for example. – mcandril Oct 28 '11 at 12:27

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