Entire function assume real values for $z=x^2+ix$

Question

Let $f$ be an entire function such that $f(z)$ is real for $z=x^2+ix$. Is there exists such a function which is not constant?

Previously I thought that $\displaystyle\int_{0}^{z}\left(\sqrt{t-1/4}-i\right) dt \ $ would work but I realized that this function isn't continous so I cannot guarantee that this integral is entire.

Answer 1

So, to reach at this answer I saw some facts:

1)$\cos(\sqrt{z})$ is an entire function - we can see this intuitively by the fact that cosine is even and with this we avoid the problem to define square root in $(-\infty,0]$ - which is a question here on Stack Exchange

2) $\cos2 \pi ix=\cosh2 \pi x$

So, to conclude I used what I think that was expected to be used, $x^2 +ix-1/4=(x+i/2)^2$

$f(z)=\cos2 \pi \sqrt{1/4-z}$ is a entire function and for $w=x^2 +ix$ we have $f(w)=\cos2 \pi \sqrt{-(x+i/2)^2}=\cos2 \pi i(x+i/2)=\cos(2 \pi ix -\pi)=-\cos2 \pi ix=-\cosh2\pi x$ which is a real number for every $x\in \mathbb{R}$

Answer 2

First we notice that the area above the given parabola is mapped eithe to the lower half plane or the upper half plane. Using conformal mapping we see that f(g) takes the upper half plane to itself with boundary to boundary. This needs not be constant for example z^3.