# How to construct a locally injective entire holomorphic function such that the modulus is axially symmetric?

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.

$$f'$$ 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 $$f(-z) = f(z)$$.