I asked this question on MSE here.
I am investigating the number of roots of the equationequations:
$$\tan(z) - z^n = 0$$ $$\sin(z)-z^n=0$$ $$\cos(z)- z^n=0$$
within the vertical strip $|\text{Re}(z)| \leq \frac{\pi}{2}$ for positive integers $n$. Numerical computations suggest interesting behavior:
$$\tan(z)-z^n=0$$
- 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):
$\tan(z)-z^5$ (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$$t_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
| $n $ | $t_n$ |
|---|---|
| 1 | 1 |
| 2 | 3 |
| 3 | 5 |
| 4 | 5 |
| 5 | 5 |
| 6 | 6 |
| 7 | 7 |
| 8 | 8 |
| 9 | 9 |
| 10 | 10 |
| 11 | 11 |
| 12 | 12 |
| 13 | 13 |
| 14 | 14 |
| 15 | 15 |
$1,3,5,5,5,6,7,8,9,10,11,12,13,14,15,16$, it seems that for $n\ge5, \ a_n=n$ but I couldn't explain this behaviour or if it is true for all $n\ge5$.
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 $|\text{Re}(z)| \leq \frac{\pi}{2}$. Also the equations $\cot(z) - z^n = 0$ and $\csc(z) - z^n = 0$ always have exactly $n+1$ roots in the strip $|\text{Re}(z)| \leq \frac{\pi}{2}$, HoweverHowever the equations $\sec(z) - z^n = 0$ is like the equation $\tan(z)-z^n$ i.e (have $n$ roots in the strip $|\text{Re}(z)| \leq \frac{\pi}{2}$ for $ n \ge 5$) .
How to explain this behavior?