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I would like to prove that if $f(z)$ is an entire function that is real if and only if $z$ is real, then $f'(z)\neq 0$ for all real $z$.

I first wrote $$f(z)=u(x,y)+iv(x,y)$$ with $z=x+iy$ and expressed our hypothesis as $v(x,y)=0 \Longleftrightarrow y=0$. So we know that $v(x,0)=0$ for all $x$. It's not clear to me where to go from here though.

I also tried to come up with examples satisfying the hypothesis: one is to take $f(z)=z$, as then $f(z)=x+iy$ and of course the hypothesis is automatically satisfied. Of course, we also have $f'(z)\neq 0$ for all real $z$. A non-zero dilation of said identity function also works of course.

Any help is much appreciated.

EDIT: I am interested in ways of proving the statement other than those already written in the MSE links in the comment.

RobPratt
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Dispersion
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    Have you tried checking if the Cauchy Riemann equations give you anything useful? – Theo Diamantakis Jun 13 '23 at 21:15
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    Would this answer your question? https://math.stackexchange.com/q/2634606/1104384 – Bruno B Jun 13 '23 at 21:26
  • Does this answer your question? Holomorphic function satisfying $f^{-1}(\Bbb R)=\Bbb R$ is of the form $f(z)=az+b$ – geetha290krm Jun 13 '23 at 23:16
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    The most direct approach is to look at local degree. i.e. for $w\in \mathbb R$ and $\gamma(t):= w +r\cdot\exp(2\pi i \cdot t)$ for $t\in [0,1]$ argue that your criterion implies $n\big(f \circ \gamma, f(w)\big)\in \big{-1,0,1\big}$ for $r\gt 0$ small enough and the winding number must be positive by the Argument Principle; conclude the function is injective in a neighborhood of $w$. – user8675309 Jun 14 '23 at 00:13
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    Hint: At a hypothetical critical point of order $k=m-1$ located at say $z_0$, the entire function $w=g(z)= f(z)- f(z_0) $ has a Taylor expansion that shows that $g(z)$ behaves much like a real multiple of $ (z-z_0)^m$. Thus $2m$ rays in the domain are locally mapped to the real axis in the $w$ plane. – MathFont Jun 14 '23 at 00:51
  • Yes I have seen those post, and I am interested in other ways of proving the statement. @MathWonk I thought of a similar idea: Let us expand $f(z)=f'(z_0)(z-z_0)+ \frac{1}{2} f''(z_0)(z-z_0)^2+O(|z-z_0|^3)$ for $z_0\in\mathbb{R}$ and assume that $f'(z_0)=0$. Because $f$ is non-constant, as you outline in your answer, there will be a term of the form $\frac{1}{m!} (z-z_0)^m$ suitably normalizing $f$ for $m\ge 2$ , and this term will map the non-real number $z=z_0+\varepsilon \exp(i\pi/m)$ to a real number, thus violating our assumptions. Would this be correct? – Dispersion Jun 14 '23 at 02:41
  • More precisely, one can argue that $w=g(z)= h(z)^m$ where (i) $h(z)$ has a real Taylor expansion and (ii) $h(z)$ is locally 1-1 and thus holomorphically invertible at $z_0$. Thus the inverse image of the real $w$ axis will be $2m$ curvilinear arcs in the $z$ plane. – MathFont Jun 14 '23 at 22:09

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