Prove that all roots of $z\tan z = k$ lie in $\Bbb R$, where $k$ is a positive, non-zero real number.

Question

The question is that given in the title;

Prove that all roots of the equation $z\tan z = k$ lie in $\Bbb R$, where $k$ is a positive non-zero real number.

I attempted a somewhat "brute force" approach by simply letting $z = a + bi$, with $a, b \in \Bbb R$. Under this assumption, if $a = 0$ (that is, $z$ is purely imaginary), we come to the conclusion that $k$ must be of the form

$$-b\tanh b = k.$$

If $b > 0$ then $\tanh b \in (0,1]$ and so $-b\tanh b < 0$. If $b < 0$ then $\tanh b \in [-1, 0)$ since $\tanh$ is odd, and so again, $-b\tanh b < 0$, and so for $k \in \Bbb R_{>0}$, we cannot have purely imaginary $z$.

Consider then the case that $z = a + bi$ where $a, b \neq 0$. Then after some manipulation, we end up with

$$z\tan z = \frac{\tan a + i\tanh b}{1 - i\tan a \tanh b}.$$

Since both $\tan$ and $\tanh$, with domain restricted to $\Bbb R$, also have codomain $\Bbb R$, this ratio must be complex, since $\tanh b = 0$ iff $b = 0$, which we excluded under assumption, and so $k$ is complex if $z$ is complex.

The only remaining possibility is that $z \in \Bbb R$, the result has been proven.

I think this is somewhat hand-wavy at best; can anyone give me a more satisfactory/elegant proof of this result?

Answer 1

This is really only a slight tweaking of your argument to remove the casework and fix a few holes. It also shows the result for $k\in\Bbb R$, not just $k\in\Bbb R^+$.

Say $z = a + bi$. Then $$ z\tan z = \frac{\tan a + i\tanh b}{1 - i\tan a\tanh b} = \left(\frac{1}{1 + \tan^2 a\tanh^2 b}\right)\left(\tan a(1 - \tanh^2 b) + i\tanh b(1 + \tan^2 a)\right). $$ $\left(\frac{1}{1 + \tan^2 a\tanh^2 b}\right)\in\Bbb R^+$, because $\tan$ and $\tanh$ are real valued when restricted to $\Bbb R$.

Hence, it is enough to show that if $\tan a(1 - \tanh^2 b) + i\tanh b(1 + \tan^2 a)$ is real, then $b = 0$. For this quantity to be real, we need $\tanh b(1 + \tan^2 a) = 0$. This forces either $\tanh b = 0$ or $1 + \tan^2 a = 0$. However, $\tan^2 a\geq 0$, so $1 + \tan^2 a\geq 1$. Thus, if $z\tan z$ is real, we must have $\tanh b = 0$. As you've said, this happens if and only if $b = 0$, which means that $a + bi\in\Bbb R$.