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Let $f$ be an entire holomorphic function, with $f'$ vanishing nowhere in the complex plane. Is it possible to construct such a function such that $|f(z)|$ is symmetric with respect both $x$-axis and $y$-axis?

My attempt is trying to find a function of the form $f(z)=ze^{g(z^2)}$, where $g$ is transcendental holomorphic and the coefficients of Taylor expansion of $g$ are real. Such function satisfies $|f(z)|=|f(-z)|=|f(\bar{z})|$, but in order that $f'$ is nowhere vanishing, we need to guarantee that $1+2z^2g'(z^2)$ has no zeros. I've no idea how to fufill this step. Maybe we cannot construct in this way, because if $1+2z^2 g'(z^2)$ has no zeros, then there exists another holomorphic function $h$ such that $1+2z^2g'(z^2)=e^{h(z)}$. This seems to be impossible.

Any ideas and comments are fully apreciated.

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$f'$ has no zeros if and only if $f' = e^h$ with an entire function $h$. It is easier to start from here and find $h$ such that $f$ has the wanted symmetry.

If $h$ is entire, real on the real axis, and an even or odd function then $f(z) = \int_0^z e^{h(w)} \, dw$ satisfies $f(\bar z) = \overline {f(z)}$ and $f(-z) = \pm f(z)$, so that $|f|$ is symmetric with respect to both x-axis and y-axis.

A simple choice is $$ f(z) = \int_0^z e^{w^2} \, dw = \sum_{n=0}^\infty \frac{z^{2n+1}}{(2n+1)n!} $$ $f'(z) =e^{z^2}$ has no zeros, also $f(\bar z) = \overline {f(z)}$ and $f(-z) = f(z)$.

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    $\begingroup$ Very nice answer! Thank you very much. $\endgroup$
    – student
    May 17 at 13:51

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