I asked this question on MSE here.

I am investigating the number of roots of the equation

$$\tan(z) - z^n = 0$$

within the vertical strip $|\text{Re}(z)| \leq \frac{\pi}{2}$ for positive integers $n$. Numerical computations suggest interesting behavior:

• The equation $\tan(z) - z = 0$ has only the trivial solution $z = 0$.
• For $n \ge 2$, the number of roots appears to depend on the value of $n$. Plots for some cases are shown below:

$\tan(z)-z$:

$\tan(z)-z^2$ (showing 3 roots):

$\tan(z)-z^3$ (showing 5 roots):

$\tan(z)-z^4$ (showing 5 roots):

$\tan(z)-z^{11}$ (showing 11 roots):

$\tan(z)-z^{12}$ (showing 12 roots):

I used this python code:

import cplot
import numpy as np
print("enter  n  ")
 n= int(input())

def f(z):
    res =np.tan(z)-z**n
    return res

plt = cplot.plot(f, (-1.6, +1.6, 1000), (-1.6, +1.6, 1000))
plt.show()

Define $a_n$ to be the number of roots that $\tan(z)-z^n$ have for $n \in \mathbb{N}$, $\frac{-\pi}{2}\le \Re(z)\le \frac{\pi}{2}$, The question is how to find $a_n$ for $n\in\mathbb{N}$?

The first few values are $1,3,5,5,5,6,7,8,9,10,11,12,13,14,15,16$, it seems that for $n\ge5, \ a_n=n$ but I couldn't explain this behaviour or if it is true for $n>12$

In contrast to $\tan(z) - z^n$, the equations $\sin(z) - z^n = 0$ and $\cos(z) - z^n = 0$ always have exactly $n$ roots in the strip $|\text{Re}(z)| \leq \frac{\pi}{2}$. How to explain this behavior?