This is an exercise from a book I was reading:
Suppose that an entire function $f:\mathbb{C} \to \mathbb {C}$ satisfies $$f(z)\in \mathbb{R} \iff z\in \mathbb{R} .$$Prove that $f'$ does not vanish on $\mathbb{R}$.
The author gave a hint; he told us to use the following theorem:
Theorem: Suppose that a function $f$ is holomorphic in an open set $U$, that $z_0$ is a point of $U$, and that $f$ takes the value $w_0$ with multiplicity $m$ at $z_0$. Then there exist $s>0$ and a domain $D$ that is contained in $U$ with the properties that
- $D$ is contained in a closed disk centered at $z_0$ that is contained in $U$,
- $f(D)=\Delta(w_0,s),$
- $f(z)\neq w_0,\ f'(z)\neq 0$ for all $z\in D-\{z_0\}$,
- for each $w$ in $\Delta ^{\ast}(w_0,s)$, there are exactly $m$ points in $D$ that are mapped to $w$ by $f$.
(Here the symbol $\Delta(w_0,s)$ denotes the open disk of radius $s$ centered at $w_0$, and $\Delta^{\ast}(w_0,s)=\Delta(w_0,s)-\{w_0\}$. )$\blacksquare$
Here's what I have tried.
My attempt: Following the author's direction, I try to use the apply the following theorem to prove the problem by contradiction.
Suppose that $f'(z_0)=0$ for some $z_0$ in $\mathbb{R}$. Choose $D$ and $s$ as in the above theorem. Fix a point $w_1$ in $\Delta(w_0,s)\cap \mathbb{R}$. According to the theorem, there are at least $2$ points in $D$ that are mapped to $w_1$ by $f$.
Now since $f$ is real on $\mathbb{R}$, so is $f'$. By choice of $D$, $f'(z)\neq 0$ for $z\in D^{\ast}:=D-\{z_0\}$. So $f'$ is either always positive or always negative on the set $A:=\{x\in\mathbb{R}\cap > D|\ x< z_0\}$. The same thing can be said about $f'$ on the set $B:=\{x\in\mathbb{R}\cap D|\ x> z_0\}$. If $f'$ happened to be positive on both of these sets, then $f$ would be strictly increasing on $A\cup \{z_0\}\cup B$, contrary to the result in the previous paragraph and our hypothesis that $f(z)$ is real iff $z$ is real. Hence $f'>0$ on $A$ and $f'<0$ on $B$, or $f'<0$ on $A$ and $f'>0$ on $B$.
I feel like I am going in the wrong direction. But I cannot find any other direction. Can anyone help me? Thanks in advance!
I will continue after the second paragraph of your attempt. Let $x_1$ and $x_2$ be points in $D$ such that $f(x_1)=f(x_2)=w_1$. Since $w_1$ is real, so are $x_1$ and $x_2$. Taking a smaller $s$ if necessary, we may assume that $(x_1,x_2)\subset D$. Since $f'$ is real on $D\cap\Bbb R$, we can apply Rolle's theorem and deduce that $f'(c)=0$ for some $c\in(x_1,x_2)$, a contradiction.