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

Question

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.

Thank you in advance!

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 $|z-\pi| \ll 1$.

Answer 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 $$\cosh (b)+\tanh (2 b m)=0$$