Prove that the equation $\tan (z)=z$ has only real roots.

Question

Prove that the equation $\tan(z)=z$ has only real roots. How to do it? The idea is that the increment of the argument need to look at the boundary of the square with a side of $\pi n$ and another that $\tan(z)-z$ has a pole at $0$. I do not know how to use it.

Answer 1

Suppose $\tan(x+iy)=x+iy$. Use the expansion $\tan(x+iy)=\frac{\sin2x}{\cos2x+\cosh2y}+i\frac{\sinh 2y}{\cos 2x+\cosh 2y}$ to get: $$\frac{1}{\cos2x+\cosh2y}(\sin2x+i \sinh2y)=x+iy .$$

This means that the 2 vectors $(\sin2x,\sinh2y)$ and $(x,y)$ are proportional, and therefore the determinant of the matrix $\begin{pmatrix}x & y \\ \sin2x &\sinh2y \end{pmatrix}=0$.

We get $x \sinh2y=y \sin2x$. Using the well-known inequalities $$|\sin t| \leq |t|,|\sinh t| \geq |t|$$ we see that we must have $x=0$ or $y=0$ (this is true because equality holds in the inequalities above iff $t=0$).

The case $y=0$ gives the wanted real solutions, and a small calculation shows that the case $x=0$ gives the one-and-only imaginary solution, namely $z=0$.

Answer 2

1. Consider Rouchet's Theorem for meromorphic functions:

• $f$, $g$ are meromorphic in open connected $G \subset \overline{\mathbb{C}}$

• $|f(z)|>|g(z)|$ on $\partial G$.

Then $$Z_{f+g} - P_{f + g} = Z_{f} - P_{f}$$ where $Z_h$, $P_h$ denote the number of zeroes and poles of $h$ inside $G$ counted with multiplicieties.

2. Now consider two functions, $f(z)=z$ and $g(z)=-\tan(z)$ on the circle $|z|=\pi n$.

As we know, $\tan(z)$ is bounded outside the $\epsilon$-neighborhood of its poles $\pi/2 + \pi n, n \in \mathbb{Z}$.

Choose $M$ s.t. the inequality $M\ge |\tan(z)|$ holds outside the $\epsilon$-neighborhood of $\pi/2 + \pi n, n \in \mathbb{Z}$.

On the circle $|z|=\pi n$ for large enough $n$ we have:

$$ |z| = \pi n > M \ge |\tan(z)| $$

since $|\pi n - (\pi/2 + \pi k)| \ge \epsilon$. (Just choose $\epsilon < \pi/2$ and adjust the corresponding $M$).

3. By Rouchet's Theorem for meromorphic functions:

$$ Z_{z-\tan(z)} - P_{z-\tan(z)} = Z_{z} - P_{z} \\ \implies Z_{z-\tan(z)} = P_{z-\tan(z)} + Z_{z} - P_{z} = 2n + 1 - 0 = 2n + 1 $$

which is exactly the number of real roots of the equation $\tan(x)=x$ on $x \in (-\pi n, \pi n)$:

• 3 roots on interval $x \in (-\pi/2, \pi/2)$
• 1 root on each interval $ x \in (-\pi k + \pi/2, -\pi (k-1) + \pi /2)$, $k=1, 2, 3, ..., n$
• 1 root on each interval $x \in (\pi/2 + \pi k, \pi /2 + \pi k)$, $k=0, 1, 2, 3,..., n-1$