I am trying to solve the following equation \begin{equation} -e^{-i2k\ell}=\frac{k-1}{k+1} \end{equation} for $k$. I thought it might be an idea to use the Lambert W function to do so, but my calculations are not really working out. Does anyone see if and maybe how this works? I would be really greatful! Thank you very much!
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$\begingroup$ Of course $k=0$ is a solution. If $l$ and $k$ are supposed to be real, then $k=0$ is the only solution. $\endgroup$– GEdgarCommented Jun 25, 2021 at 14:53
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$\begingroup$ yes! but i expect it to be infinitely many solutions... $\endgroup$– putti.123Commented Jun 25, 2021 at 15:02
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$\begingroup$ Multiply by $-1$ then since $\frac{-k+1}{k+1}$ is its own inverse, so we can compose both sides with it and get $\frac{1-e^{-i2k\ell}}{1+e^{-i2k\ell}}=k$ and then multiply by $\frac{e^{ik\ell}}{e^{ik\ell}}$ to make $\tan(k \ell) = k$. From here it's at least clear there are infinitely many solutions that are close to $k \approx \frac{\pi}{2 \ell}(2n+1)$. At this point I would probably try approximating by some other means. $\endgroup$– MerosityCommented Jun 25, 2021 at 15:46
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$\begingroup$ @Merosity how do you derive from tan$(k\ell)=k$ that $k\approx \frac{\pi}{2\ell}(2n+1)$? I am also quite confused since i thought the solutions for $k$ would be purely complex... $\endgroup$– putti.123Commented Jun 25, 2021 at 16:56
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1 Answer
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$$e^{-i2k\ell}=-\frac{k-1}{k+1}$$ Make $x=2i k$; this gives $$e^{-\ell x}=-\frac{x-2 i}{x+2 i}$$ and the solution is given in terms of the generalized Lambert function (have a look at equation $(4)$ in the paper).
This is nice from a formal point of view but not very practical. Let $k=a+ib$, cross multiply and separate the real and imaginary parts. We then need to solve $$F(a,b,\ell)=e^{2 b \ell} (b \sin (2 a \ell)+(a+1) \cos (2 a \ell))+a-1=0 \tag 1$$ $$G(a,b,\ell)=e^{2 b \ell} (b \cos (2 a \ell)-(a+1) \sin (2 a \ell))+b=0 \tag 2$$
To visualize where are (more or less precisely) the roots, make a contour plot of the function $$H(a,b,\ell)=F^2(a,b,\ell)+G^2(a,b,\ell)$$ for a given value of $\ell$.
For example, using $\ell=2.345$, there is a root close to $(1.0,-0.4)$ and Newton-Raphson method works like a charm and gives $a=0.978529$ and $b=-0.364096$.
answered Jun 26, 2021 at 14:07
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$\begingroup$ Bonjour Claude! Always a success ! $\endgroup$ Commented Jun 26, 2021 at 14:33
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$\begingroup$ @ErikSatie. Bien le bonjour, cher ami ! . Thank you $\endgroup$ Commented Jun 26, 2021 at 14:36
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$\begingroup$ @ClaudeLeibovici thank you very much! $\endgroup$ Commented Jun 27, 2021 at 15:06
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$\begingroup$ @ClaudeLeibovici I have one more question: with your substitution i end up with the solution $k=-1 -\frac{i}{2\ell} W_{e^{-2i\ell}}\left(-4i\ell e^{-2i\ell}\right)$. Here in the Lambert W function I still have $W_{e^{-2i\ell}}$ which is complex. I understand the paper in that way, that this index has to be a real number... but maybe i did something wrong? $\endgroup$ Commented Jun 27, 2021 at 17:26
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1$\begingroup$ @putti.123. You extended the generalized Lambert function to the complex domain. Congratulations ! One of these days, may be, it will bear your name. Back to serious, I do not knwo if the number of solutions is finite or not. Could take time to explore, using 3D plots or contour plots function $H$ ? Let me know. Cheers :) $\endgroup$ Commented Jun 28, 2021 at 2:10