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I am trying to solve the following equation for $k$:

\begin{equation} e^{2i\cdot l\cdot k}=\frac{k-1}{k+1} \, , \qquad l\in \mathbb{R} \end{equation}

I already know that the equation restricts the possible $k$ values to a discrete set and forces them to become complex.

What I don't understand is, if I let WolframAlpha solve the equation for different values of $l$, i get a different number of possible values for $k$. For example for $l=1$ WolframAlpha finds the solutions: \begin{align*} k_1 &\approx -0.800453-0.570033i\\ k_2 &\approx 0.800453+0.570033i \end{align*} But for $l=2$ I get \begin{align*} k_1 &\approx -0.586951-0,290409i\\ k_2 &\approx 0.586951+0,290409i\\ k_3 &\approx 1.6553+0.324277i \end{align*} If you try a lot more different values of $l$ there is always a different number of values for $k$... I don't understand why I sometimes get more solutions and sometimes less. I tried it with a lot of different values for $l$, but I wasn't able to finde a regularity which predicts the number of solutions for $k$ for different $l$.

I hope I was able to explain my problem. Thank you for your help!

putti.123
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1 Answers1

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From a completely formal point of view, this equation is simple. Let $c=2il$ to make it $$e^{-c k}=\frac {k+1}{k-1}$$ and the solution is given in terms of the generalized Lambert function (have a look at equation $(4)$).

This being said, from a practical point of view, it is not very useful. With $k=a+ib$; this gives the two equations $$F_l(a,b)=(1-a^2-b^2)+\left((a+1)^2+b^2\right) e^{-2 b l} \cos (2 a l)=0 \tag 1$$ $$G_l(a,b)=\left(a^2+2 a+b^2+1\right) \sin (2 a l)-2 b e^{2 b l}=0\tag 2$$

Now, what is interesting is to visualize if the contour line of $$\Phi_l(a,b)=\big[F_l(a,b)\big]^2+\big[G_l(a,b)\big]^2=\epsilon$$

Claude Leibovici
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