Return to Answer

Here is a sketch of an argument:

Since $\tan(z)= \frac{1}{i} \frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$, it is close to $\frac{1}{i}$ for $\Im(z)\gg 0$ and close to $-\frac{1}{i}$ for $\Im(z)\ll 0$. In particular, $|\tan(z)|$ is bounded above for large values of $\Im(z)$ and $z^n$ is bounded below for large values of $\Im(z)$, so $\tan(z)-z^n$ has no zeros outside of the region $-C\leq \Im(z)\leq C$.

We now let $\gamma$ be the contour given by the boundary of the rectangle $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$, $-C\leq \Im(z)\leq C$, minus two small disks of radius $\varepsilon$ around $\pm \frac{\pi}{2}$.

$|\tan(z)|$ stays bounded on $\gamma$, say by some $B$. For large enough $n$, we have that $|z^n|$ is strictly bigger than $B$ on all of $\gamma$, since $\gamma$ contains no $z$ with $|z|\leq 1$. So we can apply Rouche's theorem to see that $\tan(z)-z^n$ and $z^n$ have the same number of zeroes in our region, which is $n$.

This proves that in the region $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$ minus two small circles around $\pm \frac{\pi}{2}$, $\tan(z)-z^n$ has indeed exactly $n$ zeroes for large enough $n$ (where the precise numbers depend on the $\varepsilon$ you chose). However, there are also one or two zeroes in those circles: As $\tan(z)$ has a pole at $\frac{\pi}{2}$, $\tan(z)-z^n$ and $(z-\tfrac{\pi}{2})\tan(z) - (z-\tfrac{\pi}{2})z^n$ have the same zeroes. On the boundary of the circle of radius $\varepsilon$ around $\frac{\pi}{2}$, $(z-\tfrac{\pi}{2})\tan(z)\approx 1$, while $|(z-\tfrac{\pi}{2})z^n| \approx \varepsilon (\tfrac{\pi}{2})^n$, which is large for large enough $n$. So Rouche's theorem applies again to prove that $\tan(z)-z^n$ has the same number of zeroes in the interior of the circle of radius $\varepsilon$ around $\frac{\pi}{2}$ as $(z-\tfrac{\pi}{2})z^n$, which has exactly $1$. We can argue analogously for the circle of radius $\varepsilon$ around $-\frac{\pi}{2}$. We need to decide whether those two extra zeroes have real part in $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$.

Looking at real values of $z$, $\tan(z)-z^n$ is negative for $z=\frac{\pi}{2}-\varepsilon$, but $\tan(z)$ goes to $+\infty$ as $z\to \frac{\pi}{2}$ from below in the reals, so it must become positive again. So we always have a zero with real value $<\frac{\pi}{2}$ in the circle of radius $\varepsilon$ around $\frac{\pi}{2}$, which is the only one by the above argument. Around $-\frac{\pi}{2}$ we argue similarly, but because the sign of $z^n$ depends on the parity of $n$, the zero will have real value $>-\frac{\pi}{2}$ exactly if $n$ is odd (and $<-\frac{\pi}{2}$ otherwise).

Altogether we find $n+2$ zeroes in the region $-\frac{\pi}{2}\leq \Re(z)\leq \frac{\pi}{2}$ for large odd $n$, and $n+1$ for large even $n$.

Here is a sketch of an argument:

Since $\tan(z)= \frac{1}{i} \frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$, it is close to $\frac{1}{i}$ for $\Im(z)\gg 0$ and close to $-\frac{1}{i}$ for $\Im(z)\ll 0$. In particular, $|\tan(z)|$ is bounded above for large values of $\Im(z)$ and $z^n$ is bounded below for large values of $\Im(z)$, so $\tan(z)-z^n$ has no zeros outside of the region $-C\leq \Im(z)\leq C$.

We now let $\gamma$ be the contour given by the boundary of the rectangle $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$, $-C\leq \Im(z)\leq C$, minus two small disks of radius $\varepsilon$ around $\pm \frac{\pi}{2}$.

$|\tan(z)|$ stays bounded on $\gamma$, say by some $B$. For large enough $n$, we have that $|z^n|$ is strictly bigger than $B$ on all of $\gamma$, since $\gamma$ contains no $z$ with $|z|\leq 1$. So we can apply Rouche's theorem to see that $\tan(z)-z^n$ and $z^n$ have the same number of zeroes in our region, which is $n$.

This proves that in the region $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$ minus two small circles around $\pm \frac{\pi}{2}$, $\tan(z)-z^n$ has indeed exactly $n$ zeroes for large enough $n$ (where the precise numbers depend on the $\varepsilon$ you chose). However, there are also one or two zeroes in those circles: As $\tan(z)$ has a pole at $\frac{\pi}{2}$, $\tan(z)-z^n$ and $(z-\tfrac{\pi}{2})\tan(z) - (z-\tfrac{\pi}{2})z^n$ have the same zeroes. On the boundary of the circle of radius $\varepsilon$ around $\frac{\pi}{2}$, $(z-\tfrac{\pi}{2})\tan(z)\approx 1$, while $|(z-\tfrac{\pi}{2})z^n| \approx \varepsilon (\tfrac{\pi}{2})^n$, which is large for large enough $n$. So Rouche's theorem applies again to prove that $\tan(z)-z^n$ has the same number of zeroes in the interior of the circle of radius $\varepsilon$ around $\frac{\pi}{2}$ as $(z-\tfrac{\pi}{2})z^n$, which has exactly $1$. We can argue analogously for the circle of radius $\varepsilon$ around $-\frac{\pi}{2}$. We need to decide whether those two extra zeroes have real part in $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$.

Looking at real values of $z$, $\tan(z)-z^n$ is negative for $z=\frac{\pi}{2}-\varepsilon$, but $\tan(z)$ goes to $+\infty$ as $z\to \frac{\pi}{2}$ from below in the reals, so it must become positive again. So we always have a zero with real value $<\frac{\pi}{2}$ in the circle of radius $\varepsilon$ around $\frac{\pi}{2}$, which is the only one by the above argument. Around $-\frac{\pi}{2}$ we argue similarly, but because the sign of $z^n$ depends on the parity of $n$, the zero will have real value $>-\frac{\pi}{2}$ exactly if $n$ is odd (and $<-\frac{\pi}{2}$ otherwise).

Altogether we find $n+2$ zeroes in the region $-\frac{\pi}{2}\leq \Re(z)\leq \frac{\pi}{2}$ for large odd $n$, and $n+1$ for large even $n$.

EDIT: The below picture shows the minimal $n\geq 1$ for which $\tan(z)$ is dominated by $z^n$. Specifically, $n=1$ in the white region, $n=2$ yellow, $n=3$ orange, $n=4$ light red, $n=5$ dark red. So it seems that the above arguments go through for $n\geq 5$ and $\varepsilon=0.3$ (blue contour).

Here is a sketch of an argument:

Since $\tan(z)= \frac{1}{i} \frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$, it is close to $\frac{1}{i}$ for $\Im(z)\gg 0$ and close to $-\frac{1}{i}$ for $\Im(z)\ll 0$. In particular, $|\tan(z)|$ is bounded above for large values of $\Im(z)$ and $z^n$ is bounded below for large values of $\Im(z)$, so $\tan(z)-z^n$ has no zeros outside of the region $-C\leq \Im(z)\leq C$.

We now let $\gamma$ be the contour given by the boundary of the rectangle $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$, $-C\leq \Im(z)\leq C$, minus two small disks of radius $\varepsilon$ around $\pm \frac{\pi}{2}$.

$|\tan(z)|$ stays bounded on $\gamma$, say by some $B$. For large enough $n$, we have that $|z^n|$ is strictly bigger than $B$ on all of $\gamma$, since $\gamma$ contains no $z$ with $|z|\leq 1$. So we can apply Rouche's theorem to see that $\tan(z)-z^n$ and $z^n$ have the same number of zeroes in our region, which is $n$.

This proves that in the region $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$ minus two small circles around $\pm \frac{\pi}{2}$, $\tan(z)-z^n$ has indeed exactly $n$ zeroes for large enough $n$ (where the precise numbers depend on the $\varepsilon$ you chose). However, there are also one or two zeroes in those circles: As $\tan(z)$ has a pole at $\frac{\pi}{2}$, $\tan(z)-z^n$ and $(z-\tfrac{\pi}{2})\tan(z) - (z-\tfrac{\pi}{2})z^n$ have the same zeroes. On the boundary of the circle of radius $\varepsilon$ around $\frac{\pi}{2}$, $(z-\tfrac{\pi}{2})\tan(z)\approx 1$, while $|(z-\tfrac{\pi}{2})z^n| \approx \varepsilon (\tfrac{\pi}{2})^n$, which is large for large enough $n$. So Rouche's theorem applies again to prove that $\tan(z)-z^n$ has the same number of zeroes in the interior of the circle of radius $\varepsilon$ around $\frac{\pi}{2}$ as $(z-\tfrac{\pi}{2})z^n$, which has exactly $1$. We can argue analogously for the circle of radius $\varepsilon$ around $-\frac{\pi}{2}$. We need to decide whether those two extra zeroes have real part in $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$.

Looking at real values of $z$, $\tan(z)-z^n$ is negative for $z=\frac{\pi}{2}-\varepsilon$, but $\tan(z)$ goes to $+\infty$ as $z\to \frac{\pi}{2}$ from below in the reals, so it must become positive again. So we always have a zero with real value $<\frac{\pi}{2}$ in the circle of radius $\varepsilon$ around $\frac{\pi}{2}$, which is the only one by the above argument. Around $-\frac{\pi}{2}$ we argue similarly, but because the sign of $z^n$ depends on the parity of $n$, the zero will have real value $>-\frac{\pi}{2}$ exactly if $n$ is odd (and $<\frac{\pi}{2}$ otherwise).

Altogether we find $n+2$ zeroes in the region $-\frac{\pi}{2}\leq \Re(z)\leq \frac{\pi}{2}$ for large odd $n$, and $n+1$ for large even $n$.

Here is a sketch of an argument:

Since $\tan(z)= \frac{1}{i} \frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$, it is close to $\frac{1}{i}$ for $\Im(z)\gg 0$ and close to $-\frac{1}{i}$ for $\Im(z)\ll 0$. In particular, $|\tan(z)|$ is bounded above for large values of $\Im(z)$ and $z^n$ is bounded below for large values of $\Im(z)$, so $\tan(z)-z^n$ has no zeros outside of the region $-C\leq \Im(z)\leq C$.

We now let $\gamma$ be the contour given by the boundary of the rectangle $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$, $-C\leq \Im(z)\leq C$, minus two small disks of radius $\varepsilon$ around $\pm \frac{\pi}{2}$.

$|\tan(z)|$ stays bounded on $\gamma$, say by some $B$. For large enough $n$, we have that $|z^n|$ is strictly bigger than $B$ on all of $\gamma$, since $\gamma$ contains no $z$ with $|z|\leq 1$. So we can apply Rouche's theorem to see that $\tan(z)-z^n$ and $z^n$ have the same number of zeroes in our region, which is $n$.

This proves that in the region $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$ minus two small circles around $\pm \frac{\pi}{2}$, $\tan(z)-z^n$ has indeed exactly $n$ zeroes for large enough $n$ (where the precise numbers depend on the $\varepsilon$ you chose). However, there are also one or two zeroes in those circles: As $\tan(z)$ has a pole at $\frac{\pi}{2}$, $\tan(z)-z^n$ and $(z-\tfrac{\pi}{2})\tan(z) - (z-\tfrac{\pi}{2})z^n$ have the same zeroes. On the boundary of the circle of radius $\varepsilon$ around $\frac{\pi}{2}$, $(z-\tfrac{\pi}{2})\tan(z)\approx 1$, while $|(z-\tfrac{\pi}{2})z^n| \approx \varepsilon (\tfrac{\pi}{2})^n$, which is large for large enough $n$. So Rouche's theorem applies again to prove that $\tan(z)-z^n$ has the same number of zeroes in the interior of the circle of radius $\varepsilon$ around $\frac{\pi}{2}$ as $(z-\tfrac{\pi}{2})z^n$, which has exactly $1$. We can argue analogously for the circle of radius $\varepsilon$ around $-\frac{\pi}{2}$. We need to decide whether those two extra zeroes have real part in $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$.

Looking at real values of $z$, $\tan(z)-z^n$ is negative for $z=\frac{\pi}{2}-\varepsilon$, but $\tan(z)$ goes to $+\infty$ as $z\to \frac{\pi}{2}$ from below in the reals, so it must become positive again. So we always have a zero with real value $<\frac{\pi}{2}$ in the circle of radius $\varepsilon$ around $\frac{\pi}{2}$, which is the only one by the above argument. Around $-\frac{\pi}{2}$ we argue similarly, but because the sign of $z^n$ depends on the parity of $n$, the zero will have real value $>-\frac{\pi}{2}$ exactly if $n$ is odd (and $<-\frac{\pi}{2}$ otherwise).

Altogether we find $n+2$ zeroes in the region $-\frac{\pi}{2}\leq \Re(z)\leq \frac{\pi}{2}$ for large odd $n$, and $n+1$ for large even $n$.

Here is a sketch of an argument:

Since $\tan(z)= \frac{1}{i} \frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$, it is close to $\frac{1}{i}$ for $\Im(z)\gg 0$ and close to $-\frac{1}{i}$ for $\Im(z)\ll 0$. In particular, $|\tan(z)|$ is bounded above for large values of $\Im(z)$ and $z^n$ is bounded below for large values of $\Im(z)$, so $\tan(z)-z^n$ has no zeros outside of the region $-C\leq \Im(z)\leq C$.

We now let $\gamma$ be the contour given by the boundary of the rectangle $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$, $-C\leq \Im(z)\leq C$, minus two small disks of radius $\varepsilon$ around $\pm \frac{\pi}{2}$.

$|\tan(z)|$ stays bounded on $\gamma$, say by some $B$. For large enough $n$, we have that $|z^n|$ is strictly bigger than $B$ on all of $\gamma$, since $\gamma$ contains no $z$ with $|z|\leq 1$. So we can apply Rouche's theorem to see that $\tan(z)-z^n$ and $z^n$ have the same number of zeroes in our region, which is $n$.

This proves that in the region $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$ minus two small circles around $\pm \frac{\pi}{2}$, $\tan(z)-z^n$ has indeed exactly $n$ zeroes for large enough $n$ (where the precise numbers depend on the $\varepsilon$ you chose). However, there are also one or two zeroes in those circles: As $\tan(z)-z^n$ is negative for something like $z=1.5$, but $\tan(z)$ goes to $+\infty$ as $z\to \frac{\pi}{2}$ from below in the reals, it must become positive again. So there is at least one more zero there, and for odd $n$ another one near $-\frac{\pi}{2}$ by symmetry. I believe these are the only ones by looking at the fact that $\tan(z)\approx -\frac{1}{z-\frac{\pi}{2}}$, and one can probably see this rigorously by similar tricks as above.

Here is a sketch of an argument:

Since $\tan(z)= \frac{1}{i} \frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$, it is close to $\frac{1}{i}$ for $\Im(z)\gg 0$ and close to $-\frac{1}{i}$ for $\Im(z)\ll 0$. In particular, $|\tan(z)|$ is bounded above for large values of $\Im(z)$ and $z^n$ is bounded below for large values of $\Im(z)$, so $\tan(z)-z^n$ has no zeros outside of the region $-C\leq \Im(z)\leq C$.

We now let $\gamma$ be the contour given by the boundary of the rectangle $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$, $-C\leq \Im(z)\leq C$, minus two small disks of radius $\varepsilon$ around $\pm \frac{\pi}{2}$.

$|\tan(z)|$ stays bounded on $\gamma$, say by some $B$. For large enough $n$, we have that $|z^n|$ is strictly bigger than $B$ on all of $\gamma$, since $\gamma$ contains no $z$ with $|z|\leq 1$. So we can apply Rouche's theorem to see that $\tan(z)-z^n$ and $z^n$ have the same number of zeroes in our region, which is $n$.

This proves that in the region $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$ minus two small circles around $\pm \frac{\pi}{2}$, $\tan(z)-z^n$ has indeed exactly $n$ zeroes for large enough $n$ (where the precise numbers depend on the $\varepsilon$ you chose). However, there are also one or two zeroes in those circles: As $\tan(z)$ has a pole at $\frac{\pi}{2}$, $\tan(z)-z^n$ and $(z-\tfrac{\pi}{2})\tan(z) - (z-\tfrac{\pi}{2})z^n$ have the same zeroes. On the boundary of the circle of radius $\varepsilon$ around $\frac{\pi}{2}$, $(z-\tfrac{\pi}{2})\tan(z)\approx 1$, while $|(z-\tfrac{\pi}{2})z^n| \approx \varepsilon (\tfrac{\pi}{2})^n$, which is large for large enough $n$. So Rouche's theorem applies again to prove that $\tan(z)-z^n$ has the same number of zeroes in the interior of the circle of radius $\varepsilon$ around $\frac{\pi}{2}$ as $(z-\tfrac{\pi}{2})z^n$, which has exactly $1$. We can argue analogously for the circle of radius $\varepsilon$ around $-\frac{\pi}{2}$. We need to decide whether those two extra zeroes have real part in $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$.

Looking at real values of $z$, $\tan(z)-z^n$ is negative for $z=\frac{\pi}{2}-\varepsilon$, but $\tan(z)$ goes to $+\infty$ as $z\to \frac{\pi}{2}$ from below in the reals, so it must become positive again. So we always have a zero with real value $<\frac{\pi}{2}$ in the circle of radius $\varepsilon$ around $\frac{\pi}{2}$, which is the only one by the above argument. Around $-\frac{\pi}{2}$ we argue similarly, but because the sign of $z^n$ depends on the parity of $n$, the zero will have real value $>-\frac{\pi}{2}$ exactly if $n$ is odd (and $<\frac{\pi}{2}$ otherwise).

Altogether we find $n+2$ zeroes in the region $-\frac{\pi}{2}\leq \Re(z)\leq \frac{\pi}{2}$ for large odd $n$, and $n+1$ for large even $n$.