I asked this question on MSE  [here][1].




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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]][2]

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

$\tan(z)-z^3$ (showing 5 roots):
[![\tan(z)-z^3][4]][4]
$\tan(z)-z^4$ (showing 5 roots):
[![\tan(z)-z^4 ][5]][5]

$\tan(z)-z^{12}$ (showing 12 roots):
[![\tan(z)-z^{12}][6]][6]







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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 $n>12$  



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 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?


  [1]: https://math.stackexchange.com/questions/4951421/how-many-roots-does-tanz-zn-have-for-n-in-mathbbn-frac-pi2-l
  [2]: https://i.sstatic.net/AJ2zBBp8.png
  [3]: https://i.sstatic.net/jyAyRMMF.png
  [4]: https://i.sstatic.net/r9auRbkZ.png
  [5]: https://i.sstatic.net/HlsgbROy.png
  [6]: https://i.sstatic.net/rUN8AVLk.png