# Approximately solving the transcendental equation $\tan(N z)=-i\sin(z)$

everyone!

I want to find a simple analytical formula for the solutions of the following transcendental equation: $$\tan(N z) = -i\sin(z),$$ where $$N>1$$ is an integer, $$z$$ is the complex variable restricted to the range $$0\le \text{Re}(z)\le\pi$$. I can visualize the function $$\log_{10}\left| \tan(Nz)+i\sin(z) \right|$$ on the complex plane, see it below for $$N=20$$:

Clearly, there are $$N$$ solutions in the defined range that correspond to green/blue regions. I found in some article that the solution to this equation for the case $$N \gg 1$$, $$z \ll 1$$ or $$\pi-z \ll 1$$ can be written as follows: $$z_k = \frac{k\pi}{N} - i \min(k, N-k) \frac{k \pi}{N^2}$$, they are specified by black circles.

Indeed, as one can see, these are nice approximations for the roots far enough from the $$\text{Re}(z) \sim \pi/2$$ region, but close to $$\text{Re}(z) \sim \pi/2$$ it completely fails to describe those $$2$$ roots right in the center of the picture. Can anybody give me an advice how to write a kind of perturbation theory for this case, what one has to do? I am just not that experienced at analytically writing solutions to transcendental equations, usually I just solve them numerically.

UPDATE OF 06.08.2021: So, I contacted one of the authors who wrote the original approximate solution working far away from $$\text{Re}[z]\approx \pi/2$$ point, it was obtained using the following rationale:

1. Consider $$z \ll 1$$, then the zero-order solution can be obtained by considering $$\tan(Nz)=0$$, so $$z^{(0)}=j \pi/N$$, where $$j = 1,2,\ldots,N$$.
2. Now consider the first-order approximation, in order to do that, expand $$\sin(z) \approx z$$, the solution of $$\tan(N z) \approx -i z$$ for small $$z \ll 1$$ is known approximately as $$N z^{(1)} \approx \pi j - i z^{(0)}$$, where $$j = 1,2,\ldots,N$$.
3. After that one has to replace the right hand side $$z^{(0)}$$ with it's zero order approximation from 1).
4. Repeat the same procedure for the case $$|z-\pi| \ll 1$$.

Consider the functions $$f(n,a,b)=\tan (n (a+i b))+i \sin (a+i b)$$ $$F(n,a,b)= \Re\Big[f[n, a, b]\Big]^2+\Im\Big[f[n, a, b]\Big]^2$$ For a given integer value of $$n$$, the contour plot of $$F(n,a,b)$$ shows what you observed.

For $$a=\frac \pi 2$$, we need to solve for $$b$$ $$\sinh (2 b n)+\cosh (b) \left(\cosh (2 b n)+(-1)^n\right)=0$$

For sure $$b=0$$ is a trivial solution for $$n=2m-1$$. But there is another one which is the solution of $$\cosh (b) \tanh ((2 m-1)b)+1=0$$

Assuming $$b$$ to be small, the Taylor expansion is $$g(b)=\cosh (b) \tanh ((2 m-1)b)+1=1+b (2 m-1)+O\left(b^3\right)$$ So, a first guess is $$b_0=-\frac 1{2m-1}$$ which is good since $$g(b_0)\times g''(b_0) >0$$; so, by Darboux theorem, Newton method will converge without any overshoot to the solution. For example, for $$m=5$$, $$b_0=-\frac 19$$ Newton iterates are $$\left( \begin{array}{cc} k & b_k \\ 0 & -0.111111 \\ 1 & -0.171220 \\ 2 & -0.215223 \\ 3 & -0.234746 \\ 4 & -0.237307 \\ 5 & -0.237342 \end{array} \right)$$

However, a very simple and rather accurate approximation is $${\color{red}{b_m^{(1)}\sim -\frac{1}{(2 m-1)^{2/3}}}} \tag 1$$

Another possible approximation could be obtained writing $$2m-1=-\frac{\tanh ^{-1}(\text{sech}(b))}{b}$$ Expanding the rhs as Taylor series around $$b=0$$ would lead, after reversion, to $${\color{red}{b_m^{(2)}\sim -\frac{W(2 (2 m-1))}{2 m-1}}} \tag 2$$ where $$W(.)$$ is Lambert function.

If you can use Lambert function, use $$b_m^{(2)}$$ as a starting guess for Newton method since $$g\left(b_m^{(2)}\right)\times g''\left(b_m^{(2)}\right)>0\quad \text{while} \quad g\left(b_m^{(1)}\right)\times g''\left(b_m^{(1)}\right)<0 \quad \text{if} \quad m>10$$ and moreover, it is a better estimate.

As a function of $$m$$, the first solutions $$b_m$$ are $$\left( \begin{array}{cccc} m & b_m^{(1)} & b_m^{(2)} & \text{solution} \\ 2 & -0.480750 & -0.477468 & -0.481212 \\ 3 & -0.341995 & -0.349106 & -0.350398 \\ 4 & -0.273276 & -0.280578 & -0.281200 \\ 5 & -0.231120 & -0.236988 & -0.237342 \\ 6 & -0.202180 & -0.206443 & -0.206667 \\ 7 & -0.180872 & -0.183673 & -0.183825 \\ 8 & -0.164414 & -0.165948 & -0.166058 \\ 9 & -0.151252 & -0.151704 & -0.151785 \\ 10 & -0.140442 & -0.139972 & -0.140034 \\ 11 & -0.131377 & -0.130117 & -0.130167 \\ 12 & -0.123646 & -0.121708 & -0.121747 \\ 13 & -0.116961 & -0.114436 & -0.114468 \\ 14 & -0.111111 & -0.108076 & -0.108103 \\ 15 & -0.105942 & -0.102462 & -0.102485 \\ 16 & -0.101335 & -0.097465 & -0.097484 \\ 17 & -0.097198 & -0.092984 & -0.093001 \\ 18 & -0.093459 & -0.088941 & -0.088955 \\ 19 & -0.090060 & -0.085272 & -0.085284 \\ 20 & -0.086954 & -0.081926 & -0.081936 \end{array} \right)$$

When $$m$$ is large, $$1-\cosh \left(\frac{1}{(2 m-1)^{2/3}}\right) \tanh \left(\sqrt[3]{2 m-1}\right)\sim 1-\cosh \left(\frac{1}{(2 m-1)^{2/3}}\right)$$ and, by Taylor, this is $$\sim -\frac 1{3m\sqrt[3]{16 m}} \left(1+\frac 2 {3m}+\cdots \right)$$

But, if $$n=2m$$ there is no root at all for $$\cosh (b)+\tanh (2 b m)=0$$

• Dear @Claude Leibovici, thank you for your answer! You are absolutely right: only if $N$ is odd, then there is a root with the real part being exactly $\pi/2$, but what I meant in my question is to expand the functions near $z \approx \left( \frac{\pi}{2} + \delta z' \right) + i z''$, and then, from some rationale, find approximately $\delta z', z''$. So, for even $N$ there are $2$ symmetrically shifted roots (like in the original figure), while for odd $N$ there is a single root with $\delta z'=0$. Commented Aug 2, 2021 at 11:21
• Dear @Claude Leibovici, is there any way to somehow approximately write an explicit solution to the last line of yours? In a form like $b \approx f(m)$. Commented Aug 2, 2021 at 12:25
• @Sl0wp0k3. Just wait. I am working precisely this point. Commented Aug 3, 2021 at 1:39
• Dear @Claude Leibovici, thank you for such a detailed answer! Your $b^{(2)}_m$ solution looks great! I need to somehow generalize it for even $N$ so that I could find the two roots close to $z'\approx \pi/2$. During that time, I have tried to expand in Taylor series, and considered the lowest non-vanishing order, but the obtained solution failed completely. However, I have a question: whether there are any general strategies to solve transcendental equations analytically that one can start from? Commented Aug 3, 2021 at 11:56
• @Sl0wp0k3. I did the same and guess what ? I failed ! Back to the problem, I have been amazed to be able to introduce here Lambert function (my love for 64+ years - don't tell it to my wife). Cheers :-) Commented Aug 3, 2021 at 11:59