If $f(z)$ is entire, and $f(z)$ is real iff $z$ is real, prove $f'(z)$ is not equal to $0$ for all real $z$

Question

If $f(z)$ is entire, and $f(z)$ is real iff $z$ is real, prove $f'(z) \ne 0$ for all real $z$.

Edit: Sorry, somehow my preliminary efforts didn't make it in here:

I know that $f$ cannot be constant, that's a contradiction, and that if $f$ is entire, then it is analytic and holomorphic everywhere.

I have seen (but not quite understood) proofs that $f$ can have at most one zero. I have not done "winding numbers" and it seems most proofs of this employ the use of these.

Because $f$ is entire, it is continuous and so all $z$ in the upper half plane have the same sign of their imaginary part.

My first thought was by contradiction. Assume $f'(z)=0$ for some $z_0 \in C$. Then $a_1 = 0$ in the taylor series of $f$, so then $f(z)= f(z_0) + a_2*(z-z_0)^2 + O((z-z_0)^2)$. Then I'm not sure where to go from here.

I was also thinking of setting $f = u(x,y) + iv(x,y)$, and noting that $v(x,0)=0$, but I can't see how this would get me anywhere with derivatives.

Answer 1

As you stated $f(z)$ can't be constant, otherwise, from $f(z) \in \mathbb{R} \Leftrightarrow z \in \mathbb{R}$, $f(z)$ must be a real constant which happens to be real constant for non real $z$ too.

Now, by contradiction, let's assume $$f'(z)=0, \forall z \in \mathbb{R} \tag{1}$$ which also means $$f'(0)=0 \tag{2}$$ Then, using Taylor series:

$$f(z) = a_0+\sum_{n=1}a_nz^n \tag{3}$$ $$f'(z) = a_1+\sum_{n=2}na_nz^{n-1} \overset{(2)}{\Rightarrow} \color{red}{a_1=0}$$ But then $$f'(z)= \sum_{n=2}na_nz^{n-1}= z\left(\sum_{n=2}na_nz^{n-2}\right)=z\cdot g_1(z) \overset{(1)}{\Rightarrow} g_1(z)=0, \forall z \in \mathbb{R}, z \ne 0$$ $g_1(z)$ is also entire and continuous. This means that for example $0=g_1\left(\frac{1}{n}\right)\rightarrow g_1(0)=0$, as $n \rightarrow \infty, n \in \mathbb{N}$, so $g_1(0)=0 \Rightarrow \color{red}{a_2=0}$.

But then $$f'(z)=z\left(\sum_{n=2}na_nz^{n-2}\right)=z^2\left(\sum_{n=3}na_nz^{n-3}\right)=z^2\cdot g_2(z) \overset{(1)}{\Rightarrow} g_2(z)=0, \forall z \in \mathbb{R},z \ne 0$$ similarly, $g_2(z)$ is also entire (and continuous), thus $g_2(0)=0 \Rightarrow \color{red}{a_3=0}$.

Continuing like this, by induction, we conclude $\color{red}{a_n=0, \forall n>0}$. But then, from $(3)$, $f(z)=a_0$ which is a constant, contradiction with the statement at the beginning.

Answer 2

Step 1. Clearly $\mathrm{Im}\,f(x+iy)\ne 0$, whenever $y\ne 0$. In particular, $\mathrm{Im}\,f(z)$ maintains sign in the upper half plane and the lower half plane, and it has different signs in each half plane. Without loss of generality, $\mathrm{Im}\,f(x+iy)> 0$, if $y>0$, and $\mathrm{Im}\,f(x+iy)< 0$, if $y<0$.

Step 2. Assume that $f'(x)=0$, for some $x\in\mathbb R$. Without loss of generality, assume that $x=0$. Otherwise, replace $f$ by $f-f(0)$. So $f'(0)=0$, and $f\not\equiv 0$, implies the existence of a $k\in\mathbb N$, such that $f(z)=z^kg(z)$, where $g$ entire and $g(0)\ne 0$. Thus, there exists an $a\in\mathbb R$, such that, if $f=u+iv$, then $$ 0\ne a=f^{(k)}(0)=\partial_x^kf(0)=\partial_x^ku(0)=\partial_x^ku(0)=\partial_y^kv(0). $$ Clearly $k$ is odd, otherwise $v$ would have a local minimum, which is impossible, since $v<0$, in the lower half plane. In fact, $a>0$, since $v$ is increasing in $y$ at $y=0$. Hence $f(x)>0$ and $f$ strictly increasing for $x>0$, and $x$ sufficiently small and $f(x)<0$ and $f$ also strictly increasing, for $x<0$ and $|x|$ sufficiently small. But this means, that the equation $$ f(x+iy)=\eta $$ would have $k$ solution for $|\eta|$ is a small disk centred at $0$. But all the $k$ roots have to be negative, which means that $k=1$, since $f$ is strictly increasing. Hence, $f'(0)\ne 0$.

In you need to show that $f$ is linear, then, observe that

a. $f$ has to be strictly increasing, in $x\in\mathbb R$, and by Picard's Little Theorem, $f$ is a polynomial, since, the neighbourhoods of infinity do not contain absolutely small real numbers. But the only root in real, and there is only one real real root and single, and hence $f$ is a linear function of the form $$ f(z)=az+b $$ where $a,b\in\mathbb R$ and $a\ne 0$.